Unformatted text preview: NOTES ON INEQUALITIES
FELIX LAZEBNIK Order and inequalities are fundamental notions of modern mathematics. Calculus and Analysis depend heavily on them, and properties of inequalities provide the
main tool for developing these subjects. Often students are not sure what they are
allowed to take for granted when they argue about inequalities. These brief notes
are intended to help them to review the basics, improve their skills in working with
inequalities, and present two inequalities which have many applications. Here is
the list of sections. Content:
1.
2.
3.
4.
5.
6.
7.
8. Order on R and Basic Properties of Inequalities
Solving Inequalities: Case analysis
Solving Inequalities: Method of Intervals
Proving Inequalities by Induction
Jensen’s Inequality
The ArithmeticGeometric Mean Inequality (AGM)
Problems
Hints and Answers to Problems 1. Order on R and Basic Properties of Inequalities
We assume that the reader is well familiar with real numbers (or just reals), with
the algebraic properties of operations on them, and with basic properties of their
ordering. We denote the set of all real numbers by R, the set of all negative reals
by R− , and the set of positive reals by R+ . Then R− , { 0 }, and R+ partition R,
which is another way of saying that every real number is either negative, or zero,
or positive, and no real number has two of these properties. We take for granted
that the following properties hold.
•
•
•
•
•
•
•
• the sum of any two positive reals is positive
the product of two positive reals is positive
the sum of any two negative reals is negative
the product of any two negative reals is positive
the product of a positive real and a negative real is negative
number 1 is positive
x positive (negative) if and only if x−1 is positive (negative)
the ratio of a positive and a negative numbers is negative Date : December 30, 2009.
1 2 FELIX LAZEBNIK • for any two reals x, y , xy = 0 if and only if x = 0 or y = 0
• for any two reals x, y , x < y if and only if y − x is positive.
• for any positive (negative) real x, 0 < x (x < 0).
For convenience, we introduce another symbol >, called greater, and we write
x>y if y < x. The abbreviation for (x < y ) ∨ (x = y ) is x ≤ y . If x ≤ y , we say that x is at
most y , or, equivalently, y is at least x. Similarly, for x ≥ y . Hence, 3 ≤ 3 and
3 ≤ 5 are true. If 0 ≤ x or, equivalently, x ≥ 0, we say that x is nonnegative,
and if x ≤ 0 or, equivalently, 0 ≥ x, we say that a is nonpositive.
The following property is used often, and we wish to state it separately.
• for every x, x2 ≥ 0, and x2 = 0 if and only if x = 0.
Moreover, the sum of n ≥ 2 nonnegative (nonpositive) numbers is nonnegative (nonpositive), and it is equal to 0 if and only if all addends are
equal to 0.
Now we want to (ﬁnally) prove some more subtle properties which relate the
inequalities on R and operations on R. (So, please forget that you are familiar with
them!)
Theorem 1.1.
(1) For all real x, y, z , if x < y and y < z then x < z .
(2) For all real x, y, z , x < y if and only if x + z < y + z .
(3) For all real x, y and positive z , x < y if and only if xz < yz .
(4) For all real x, y and negative z , x < y if and only if xz > yz .
(5) For all real a, b, x, y , if a < b and x < y , then a + x < b + y .
Similar statement holds for the sum of any n ≥ 2 inequalities.
(6) For all positive a, b, x, y , if a < b and x < y , then ax < by .
Similar statement holds for the product of any n ≥ 2 inequalities of
positive numbers.
(7) For all integer n ≥ 1, and any positive a, b, a < b if and only if an < bn .
Proof. (1) By the deﬁnition of <: if x < y , then y − x is positive, and if y < z , then
z − y is positive. The sum of positive numbers is positive (a part of the deﬁnition
of positive reals). So (y − x) + (z − y ) = z − x is positive. By the deﬁnition of <,
x < z.
(2) (y + z ) − (x + z ) = y − x. Therefore the diﬀerence on the left is positive if
and only if the diﬀerence on the right is positive. This means that the inequalities
imply each other.
(3) yz − xz = (y − x)z . By the deﬁnition of <: if x < y , then y − x is positive.
Since z is positive (given), and the product of positive numbers is positive (part of
the deﬁnition of positive reals), (y − x)z is positive. Hence yz − xz is positive, and
so xz < yz .
(4) Left for the reader.
(5) Proof 1. By (2), a < b implies a + x < b + x, and x + b < y + b. As addition
of reals is commutative, the last inequality can be written as b + x < b + y . Then, NOTES ON INEQUALITIES 3 by transitivity of inequalities (Property (1)), a + x < b + x and b + x < b + y imply
a + x < b + y.
Proof 2. We wish to show that a + x < b + y . By the deﬁnition of <, this amounts
to showing that (b + y ) − (a + x) > 0.
Note that (b + y ) − (a + x) = (b − a) + (y − x). As b − a > 0 (since a < b)
and y − x > 0 (since x < y ), and the sum of positive reals is positive, we have
(b − a) + (y − x) > 0. Hence, (b + y ) − (a + x) > 0, and the proof is ﬁnished.
Let us generalize (5): For all n ≥ 2, the inequalities ai < bi , i = 1, . . . , n, imply
a1 + . . . + an < b1 + . . . + bn .
We prove this by the method of mathematical induction (on n). The base case,
n = 2, has been established in (5): a1 = a, b1 = b, a2 = x, b2 = y . Suppose the
statement holds for any n = k ≥ 2 inequalities. We have to show that it holds for
any n = k + 1 inequalities.
Let xi < yi for all i = 1, . . . , k + 1. We have to show that
x1 + . . . + xk + xk+1 < y1 + . . . + yk + yk+1 .
Note that x1 + . . . + xk < y1 + . . . + yk by the induction hypothesis, and xk+1 < yk+1
as given. These two inequalities can be added, as was shown in the base case
(n = 2). By doing this, we obtain
(x1 + . . . + xk ) + xk+1 < (y1 + . . . + yk ) + yk+1 .
When several real numbers are added, the parentheses can be placed in arbitrary
way due to the associative property of addition of reals. Hence, the statement holds
for any for n = k + 1 inequalities, and the proof is ﬁnished.
By now the reader should understand the logic of such proofs, and we proceed by
omitting similar simple reasons in our explanations.
(6) Proof 1. By (3), a < b and x > 0 imply ax < bx. Similarly, x < y and
b > 0 imply xb < yb. As multiplication or reals is commutative, the last inequality
is equivalent to bx < by . By transitivity of <, ax < bx and bx > by imply ax < by .
Proof 2. by − ax = by − bx + bx − ax = b(y − x) + (b − a)x. As all b, y − x, b − a, x
are positive, by − ax is positive. Hence, ax < by .
We leave the proof of the generalization of this property (by the method of
mathematical induction on n) to the reader. (7) If we have established generalization of (6) for any n ≥ 2 inequalities with
positive sides, then the result follows from multiplying n same inequalities a < b,
where a, b > 0. If the generalization of (6) has not been established, the result can
be proven by induction on n.
Example 1. Let 0 < a < b. Then for any integer n ≥ 2, √
n a< √
n b. 4 FELIX LAZEBNIK Proof. We prove it by assuming the contrary and obtaining a contradiction. Sup√
√
√
√
pose n a > n b. √
Since both sides are positive, ( n a)n > ( n b)n by (7). As
√n
( n a) = a and ( n a)n = b, this is equivalent to a > b, a contradiction with
√
√
the assumption that a < b. Hence, n a < n b.
The absolute value of a real number x, denoted by x, is deﬁned as x for x ≥ 0,
and as −x for x < 0. x can be thought as the distance from the point on the
real line corresponding to number x to the origin. More generally, for two points
A(a) and B (b) on the real line, the distance AB = a − b. Important properties of
absolute value are the following: for all x, y
x x
xy  = x y , =
(y = 0), x + y  ≤ x + y .
y
y 
These properties of the absolute value function follow from the deﬁnition of operations with reals and their geometric interpretation as points of a line. 2. Solving inequalities: Case analysis
In the examples below we use symbols ∨ and ∧ to denote the disjunction and
conjunction of predicates (i.e., open sentences, equations, inequalities, . . .). If the
reader is unfamiliar with them, the usage of connectives “or” and “and”, respectively, does the same.
Example 2. Find all real solutions of the following inequalities.
(i) x2 − 3x + 2 > 0 (ii) 3x + 2
≤1
x (iii) x2 − x − 1
< 3.
x+2 Solution. (i)
x2 − 3x + 2 > 0 ⇔
(x − 1)(x − 2) > 0 ⇔
(x − 1 > 0) ∧ (x − 2 > 0)
(x − 1 < 0) ∧ (x − 2 < 0) ⇔
(x > 1) ∧ (x > 2)
(x < 1) ∧ (x < 2) ⇔
(x > 2) ∨ (x < 1). The solution set is (−∞, 1) ∪ (2, ∞).
(ii) 3x + 2
≤1⇔
x
(x > 0) ∧ (3x + 2 ≤ x)
(x < 0) ∧ (3x + 2 ≥ x) ⇔
[(x > 0) ∧ (x ≤ −1]
[(x < 0) ∧ (x ≥ −1)] ⇔ The solution set is (−1, 0). (x ∈ ∅) ∨ (−1 ≤ x < 0). NOTES ON INEQUALITIES 5 (iii) x2 − x − 1
<3⇔
x+2
(x + 2 > 0) ∧ (x2 − x − 1 < 3(x + 2))
(x + 2 < 0) ∧ (x2 − x − 1 > 3(x + 2)) ⇔
(x > −2) ∧ (x2 − 4x − 7 < 0)
(x < −2) ∧ (x2 − 4x − 7 > 0) ⇔
√
√
√
√
(x > −2)∧(2− 11 < x < 2+ 11)
(x < −2)∧[(x < 2− 11)∨(x > 2+ 11)] ⇔
√
√
√
(−2 < x < 2+ 11)
[(x < −2) ∧ (x < 2 − 11)] ∨ [(x < −2) ∧ (x > 2+ 11)] ⇔
(−2 < x < 2 + √ 11) [(x < 2 − √ 11) ∨ (x ∈ ∅)] ⇔ √
√
(−2 < x < 2 + 11) ∨ (x < 2 − 11).
√
√
The solution set is (−∞, 2 − 11) ∪ (−2, 2 + 11).
The method we used above is often referred as case analysis. It boils to partitioning the set of all possible values of x into disjoint subsets, and considering
separately each of them. The following examples provide additional illustrations of
the method.
Example 3. Find all real solutions of the inequality
x − 2
≤ 2.
x+1 Solution. As a ≤ b ⇔ −b ≤ a ≤ b ⇔ (−b ≤ a) ∧ (a ≤ b), we have:
x − 2
≤2⇔
x+1 x−2
≤2⇔
x+1
[(x+1 > 0) ∧ (−2x−2 ≤ x−2 ≤ 2x+2)]
[(x+1 < 0) ∧ (−2x−2 ≥ x−2 ≥ 2x+2)] ⇔
−2 ≤ [(x > −1) ∧ (x ≥ 0) ∧ (x ≥ −4)] [(x < −1) ∧ (x ≤ 0) ∧ (x ≤ −4)] ⇔ (x ≥ 0) ∨ (x ≤ −4).
The solution set is (−∞, −4] ∪ [0, ∞) . 6 FELIX LAZEBNIK Example 4. Find all real solutions of the inequality
x + 1 + x + 3 + x − 5 ≤ 8. Solution. In order to rewrite the inequality without absolute value signs, we
divide reals into four intervals deﬁned by the points where the expressions under
the absolute value signs change their signs, namely by x = −3, x = −1, and x = 5.
Then we have:
x + 1 + x + 3 + x − 5 ≥ 8 ⇔
[(x ≤ −3) ∧ (−(x + 1) + (−(x + 3)) + (−(x − 5)) ≤ 8]
[(−3 < x ≤ −1) ∧ (−(x + 1) + (x + 3) + (−(x − 5)) ≤ 8]
[(−1 < x ≤ 5) ∧ ((x + 1) + (x + 3) + (−(x − 5)) ≤ 8]
[(5 < x) ∧ ((x + 1) + (x + 3) + (x − 5) ≤ 8]
[(−3 < x ≤ −1) ∧ (−x + 7 ≤ 8]
[(−1 < x ≤ 5) ∧ (x + 9 ≤ 8)]
[(5 < x) ∧ (3x − 1 ≤ 8] ⇔ [(x ≤ −3) ∧ (−3x + 1 ≤ 8] ⇔
[(−3 < x ≤ −1) ∧ (x ≥ −1]
[(−1 < x ≤ 5) ∧ (x ≤ −1)]
[(5 < x) ∧ (x ≤ 3] ⇔ [(x ≤ −3) ∧ (x ≥ −7/3] (x ∈ ∅) ∨ (x = −1) ∨ (x ∈ ∅) ∨ (x ∈ ∅) ⇔ x = −1. The solution set is {−1}.
The answer suggests the existence of a simpler solution. Interpreting a − b in
geometric terms as the distance on the real line between points A(a) and B (b),
we are looking at points X (x) such that the sum of distances from X to points
corresponding to −3, −1 and 5 is at least 8. Note that 8 is the distance between
points corresponding −3 and 5. Therefore x + 3 + x − 5 = 8 for all x in [−3, 5],
and x + 3 + x − 5 > 8 for all x not in [−3, 5]. If x = −1, then x + 1 > 0, and
x + 1 + x + 3 + x − 5 > 8. Hence, x = −1, is the only value of x such that and
x +1 + x +3 + x − 5 ≤ 8.
Example 5. Solve: x8 − x5 + x2 − x + 1 ≤ 0. Solution. Trying to factor the polynomial in the right hand side seems like a hard
problem. Experimenting with diﬀerent values of x, we fail to ﬁnd any solutions.
Therefore we will try to prove that no solutions exists, or in other words, that
f (x) = x8 − x5 + x2 − x + 1 > 0 for all x. Trying partitioning all reals into several
subsets such that the statement can be easily veriﬁed in each of them, one may
ﬁnally arrive to the following three cases: x ≤ 0, 0 < x < 1, and x ≥ 1. Case 1: x ≤ 0. For such x, f (x) is the sum of four nonnegative terms x8 , −x5 , x2 , −x,
and 1. Hence f (x) > 0 in this case.
Case 2: 0 < x ≤ 1. We rewrite f (x) in the following way: f (x) = x8 + x2 (1 −
x ) + (1 − x). As 0 < x ≤ 1, 1 − x3 ≥ 0 and 1 − x > 0. Hence f (x) is the sum
3 NOTES ON INEQUALITIES 7 of two positive numbers x8 , 1 − x, and one nonnegative number x2 (1 − x3 ). Hence
f (x) > 0 in this case.
Case 3: x ≤ 1. We rewrite f (x) in the following way: f (x) = (x8 − x5 ) + (x2 −
x) + 1 = x5 (x3 − 1) + x(x − 1) + 1. As x ≥ 1, both x3 − 1 and x − 1 are nonnegative.
Hence f (x) is the sum of two nonnegative numbers x5 (x3 − 1) and + x(x − 1), and
a positive number 1. Hence f (x) > 0 in this case.
Therefore the solution set of our inequality is the empty set ∅. 3. Solving Inequalities: Method of Intervals
There exists another way of solving inequalities, which uses continuity of functions and the Intermediate Value Theorem (IVT). Though the justiﬁcation of this
approach is based on calculus, the Method of Intervals, which we state further
below, can be easily understood and used without any knowledge of calculus.
We remind the reader that a function f is continuous on [a, b] if limx→c f (x) =
f (c) for all c ∈ (a, b), limx→a+ f (x) = f (a), and limx→b− f (x) = f (b). Intuitively,
f is continuous if its graph can be drawn without taking the pen oﬀ the paper. It
can be shown that all polynomials are continuous functions on R, and all rational
functions are continuous at the points they are deﬁned. More general, all elementary
functions are continuous in their domain. Elementary functions include polynomial,
rational functions, power functions, exponential functions, trigonometric functions,
and inverses of all these functions where they exist. Absolute value function is
another example of a continuous function. It can be proven that the sum, diﬀerence,
product and ratio of continuous functions is continuous (in all points of its domain).
Similarly, the compositions of continuous functions are also continuous.
We also remind the reader the IVT.
Theorem 3.1. ( Intermediate Value Theorem)
A continuous function f : [a, b] → R has the property that if f (a) and f (b) are
distinct, then for every number M between f (a) and f (b), there exists c ∈ (a, b)
such that f (c) = M .
In particular, when one of f (a), f (b) is positive and another negative, and M = 0,
there exists c ∈ (a, b) such that f (c) = 0. An immediate corollary from the IVP is
the following statement:
Corollary 3.2. Let f : [a, b] → R be continuous function and let z1 , . . . , zn , z1 <
z2 < . . . < zn , be all distinct solutions of the equation f (z ) = 0 in (a, b). Then on
each of the n + 1 intervals (a, z1 ), (zi , zi+1 ), and (zn , b), all values of f are of the
same sign, i.e., are all positive or all negative.
The validity of this corollary should be clear: if at least one of these intervals
contains two points such that f is positive at one of them and negative at another,
by the IVT there would be a point between them (and so inside the interval) where
f would take value zero. But there are no such points in the interval because zi
are the only points on (a, b) where f is zero. Hence, the sign of all values of f on
(zi , zi+1 ) can be determined by its sign at one point of the interval. 8 FELIX LAZEBNIK A method of solving inequalities based on this corollary is often referred to as
the Method of Intervals. Suppose we have to solve an inequality
f (x) > 0,
where f is a function continuous on an interval (a, b), except, maybe, of a ﬁnite
number of (discontinuity) points. We allow a = −∞ or b = ∞.
Consider the equation f (x) = 0.
Suppose it has ﬁnitely many solutions on (a, b). Let x1 , x2 , . . . , xk represent all
points of (a, b) where f is not continuous, together with all solutions of the equation
f (x) = 0. Assume
x1 < x2 < . . . < xk .
These k points divide the interval into k + 1 subintervals:
(a, x1 ), (x1 , x2 ), (x2 , x3 ), . . . , (xk , b).
On each of these intervals, f is continuous and nonzero. Therefore, in order to
determine the sign of f on each interval, we just chose a point from it and determine
the sign of f at this point. By the Corollary 3.2, for all points of the interval, this
sign will be the same.
Example 6. Solve the inequality
x2 − x − 6
> 0.
x2 − 1
2 x
Solution. Let f (x) = x x−−−6 . Being a rational function, f is continuous at all
21
points where it is deﬁned, i.e., for all real numbers, except values of x which make
x2 − 1 equal to zero. As x2 − 1 = (x − 1)(x + 1), f is not continuous at x = −1 and
x = 1.
The equation f (x) = 0 is equivalent to x2 − x − 6 = 0 ∧ (x = ±1). As
2
x − x − 6 = (x − 3)(x + 2), f (x) = 0 for x = −2 and x = 3.
Therefore we consider the following ﬁve intervals: (−∞, −2), (−2, −1), (−1, 1), (1, 3), (3, ∞). Choosing a point in each of them, and deciding whether f is positive or negative
at this point, we obtain
f (−100) > 0, f (−1.5) < 0, f (0) > 0, f (2) > 0, f (100) > 0.
Therefore, f (x) > 0 precisely on (−∞, −2), (−1, 1), and (3, ∞). Before proceeding to other examples, we wish to make several simple comments.
• It is clear that the inequality f (x) < 0 can be solved in a similar way.
• An inequality of the form f (x) > g (x) (f (x) < g (x)) is equivalent to h(x) >
0 (h(x) < 0), where h(x) = f (x) − g (x). Instead of introducing h, one can
just solve the equation f (x) = g (x), and proceed as before by testing a
number from each obtained interval. NOTES ON INEQUALITIES 9 • In case we have f (x) ≥ 0, or f (x) ≤ 0, or f (x) ≥ g (x), or f (x) ≤ g (x),
we have to add the solutions of the corresponding equation to the points of
some intervals.
2
x
For example, the solution of x x−−−6 ≥ 0 would consist of all points of
21
the intervals
(−∞, −2], (−1, 1), and [3, ∞). • In many cases, in order to determine the sign of a function at a point, one
does not have to compute the exact value of the function.
Example 7. Find all real solutions of the inequality
x2 − 7x + 1 < 5. Solution. First we consider the corresponding equation.
x2 − 7x + 1 = 5 ⇔ [(x2 − 7x + 1 = 5) ∨ (x2 − 7x + 1 = −5)] ⇔
(x2 − 7x − 4 = 0) ∨ (x2 − 7x + 6 = 0)] ⇔
[(x = (7 − √ 65)/2) ∨ (x = (7 + √ 65)/2)] ∨ [(x = 1) ∨ (x = 6)]. We consider ﬁve intervals:
√
√
√
√
(−∞, (7 − 65)/2), ((7 − 65)/2, 1), (1, 6), (6, (7 + 65)/2), ((7 + 65)/2, ∞).
Let f (x) = x2 − 7x + 1 − 5. As f (−100) > 0, f (0) < 0, f (2) > 0, f (7) < 0, and f (100) > 0,
7 − √65 7 + √65
we conclude that the solution set of the inequality is
,1
6,
.
2
2
4. Proving Inequalities by Induction.
The method of mathematical induction is often very eﬀective for proving inequalities. We remind ourselves that there are two most often used versions of the
method. We refer to the ﬁrst one as to “mathematical induction”, and the second
– as to “strong mathematical induction”. The adjective “mathematical” is used to
stress the fact that we mean a particular mathematical statement, since the word
“induction“ is applied to a much broader notion.1
Let n0 be a ﬁxed integer, and suppose we wish to prove that for all integers
n ≥ n0 , and P (n) be a statement referring to n. We remind the reader2 that
according to the method of Mathematical Induction , we prove the statement
“for all n, n ≥ n0 , P (n)”
1See, e.g. http://en.wikipedia.org/wiki/Inductive reasoning
−
2We assume that the reader has had some experience with mathematical induction. 10 FELIX LAZEBNIK if we establish the following two facts:
(i) P (n0 ) is correct (the base case);
(ii) if P (k ) is true, then P (k + 1) is true for each k ≥ n0 .
The method of Strong Mathematical Induction asserts that we prove the
statement
“for all n, n ≥ n0 , P (n)”
if we establish the following two facts:
(i) P (n0 ) is correct (the base case);
(ii) if P (k ) is true for all n0 ≤ k < n, then P (n) is true.
At the ﬁrst glance, the second method seems to be more convenient to use than
the ﬁrst, since we are allowed to assume more. This is true. Nevertheless, one can
show that the methods are equivalent.
Let us present several examples of proofs of inequalities by mathematical induction. In all our examples we will apply the following logic. Suppose we have A > B ,
and we wish to show that A > C . Then, if we show that B ≥ C , we are done. If
B ≥ C is false, or if we are unable to show this, then A still can be greater than C ,
but it has to be established in some other way.
Example 8. Prove that for all integers n, n ≥ 2,
1
1
2n
1 + + ... + >
2
n
n+1
Solution. We use the method of Mathematical Induction.
Let n = 2. We have to show that 1 + 1 > 4 . As 1 + 1/2 = 3/2, and 3/2 > 4/3
2
3
(3 = 9 > 2 · 4 = 8), the base case is established.
2 Suppose k ≥ 2 and the statement is true for n = k , i.e.,
1
1
2k
1 + + ... + >
.
2
k
k+1
We wish to show that the statement is true for n = k + 1, i.e.,
1
1
2(k + 1)
2(k + 1)
1 + + ... +
>
=
.
2
k+1
(k + 1) + 1
k+2 (1) (2) From (1), we have:
1
1
1
2k
1
1 + + ... +
+
>
+
.
(3)
2
k
k+1
k+1 k+1
As the expressions in the left hand sides of (2) ) and (3) are equal, it is suﬃcient
to show that
2k
1
2(k + 1)
+
≥
(4)
k+1 k+1
k+2
The last inequality is equivalent to
2k + 1
2k + 2
≥
⇔ (2k + 1)(k + 2) ≥ (k + 1)(2k + 2) ⇔
k+1
k+2
2k 2 + 5k + 2 ≥ 2k 2 + 4k + 2 ⇔ k ≥ 0. NOTES ON INEQUALITIES 11 As k ≥ 2, (4) holds, and the proof is ﬁnished.
To see how this solution reﬂects on our general scheme of inductive proofs dis1
cussed above, take n0 = 2, P (n) to be the statement 1 + 1 + . . . + n > n2n , P (k )
2
+1
1
to be the statement (2), P (k + 1) to be the statement (3), A = 1 + 1 + . . . + k+1 ,
2
B= 2k
k+1 + 1
k+1 , and C = 2(k+1)
k+2 . Example 9. Prove that for all integers n, n ≥ 1,
2n−1 ≤ n! Solution. We use the method of Mathematical Induction.
Let n = 1. We have to show that 20 ≥ 1!. As 20 = 1! = 1, the base case is
established.
Suppose k ≥ 2 and the statement is true for n = k , i.e.,
2k−1 ≤ k ! (5) 2(k+1)−1 ≤ (k + 1)! ⇔ 2k ≥ (k + 1)! (6) 2k = 2 · 2k−1 ≤ 2 · k ! (7) We wish to show that the statement is true for n = k + 1, i.e.,
From (5), we have: As the expressions in the left hand sides of (6) ) and (7) are equal, it is suﬃcient
to show that
2 · k ! ≤ (k + 1)!
(8)
The last inequality is equivalent to 2≤k+1 As k ≥ 1, (8) holds, and the proof is ﬁnished.
Note that the inequality in Example 9 could be established faster, by multiplying
obvious n inequalities:
1 ≤ 1, 2 ≤ 2, 2 ≤ 3, 2 ≤ 4, . . . 2 ≤ n. Solution of many problems suggested at the end follow the patterns of our solutions to Examples 8 and 9.
The following example illustrates how one can use the Strong Mathematical
Induction for proving inequalities.
Example 10. Let the sequence {an }n≥1 be deﬁned as follows: a1 = 1, a2 = 2,
and for all n ≥ 3, an = an−1 + an−2 . Here are several ﬁrst successive terms of the
sequence:
1, 2, 3, 5, 8, 13, 21, 34, 55, . . .
Prove that for all n, n ≥ 1, an < 1.7n . 12 FELIX LAZEBNIK Solution. We use the method of Strong Mathematical Induction.
Let us check the statement for the ﬁrst two values of n, n = 1 and n = 2. We
have to show that a1 < 1.71 , and a2 < (1.7)2 . This is obvious, as 1 < 1.7, and
2 < 1.72 = 2.89.
It establishes the base case. Why we need to check the statement for n = 2 will
become clear later.
Suppose the statement is true for all k , 1 ≤ k < n, i.e.,
ak ≤ 1.7k . (9) an < 1.7n . (10) We wish to show that
From the deﬁnition of the sequence and the induction hypothesis, we have:
an = an−1 + an−2 < 1.7n−1 + 1.7n−2 . (11) Comparing (10) ) and (11), we conclude that in order to prove (10) ), it is suﬃcient
to show that
1.7n−1 + 1.7n−2 < 1.7n .
Dividing both sides of the last inequality by 1.7
an equivalent inequality n−1 (12) (which is positive), we obtain 1 + 1.7 < 1.72
As 2.7 < 2.89, (10) holds for all n ≥ 3, and the proof is ﬁnished.
Question: Let an be as in the last example. Can an < αn still holds for all n ≥ 1
for some α < 1.7? What is the least α such that an < αn holds for all n ≥ 1? 5. Jensen’s Inequality
A function f : (a, b) → R is called convex up (same as concave up) on (a, b), if
for all x1 , x2 ∈ (a, b),
x1 + x2
f (x1 ) + f (x2 )
f
≤
.
(13)
2
2 Changing the sign in the inequality above to ≥, we obtain the deﬁnition of a function
convex down (same as concave down) on (a, b).3 If f is convex up (down) on (a, b),
then its graph lies below (above) the line passing through any of its two points, as
illustrated below.
3To remember the shape of a convex up function, think about the shape of letter “U” in “Up”. NOTES ON INEQUALITIES 13 y y=f(x) y=f(x)
Convex up Convex down x
It turns out that for a convex up (down) function, an analog of the inequality
(13) also holds for any n ≥ 2 numbers from the interval. The following inequality
was published by J.L. Jensen (1859  1925) in 1906.
Theorem 5.1. (Jensen’s Inequality) Let f be a function convex up on (a, b).
Then for any n ≥ 2 numbers xi ∈ (a, b),
n
n
f (xi )
i=1 xi
f
≤ i=1
,
n
n
and that the equality is attained if and only if f is linear or all xi are equal.
For a convex down function, the sign of the inequality changes to ≥. Proof. The proof we present uses the idea of ‘up’ and ‘down’ induction (having
nothing to do with ‘up’ and ‘down’ convexity) which is credited to A.L. Cauchy
(1789  1857).
The idea of ‘up’ and ‘down’ induction is the following. After establishing the
base case (for n = n0 ), one then proves the statement for a particular inﬁnite
increasing sequence of integers. The sequence can be, e.g., n = 2k , k ≥ 1, and a
proof that the statement holds for these n can use an independent inductive (with
respect to k ) argument. Once it is done, one shows that if a statement is true for
n = k > n0 , then it must me true for n = k − 1. After it is done, one concludes
that the statement is true for all the integers n, n ≥ n0 .
Meditating a minute over this approach, we conclude that it is as valid as one of
the more traditional versions of mathematical induction.
In this proof we ﬁrst use induction on k to show that the inequality is satisﬁed
for all n = 2k (the ‘up’ part). The statement is correct for k = 1, as it states the
given fact f is convex up on (a, b). To illustrate the main step (the passage from
n = 2k to n = 2k+1 ), we ﬁrst consider next two values of n, namely n = 22 = 4 and
n = 23 = 8.
Suppose n = 22 = 4. Then
x1 +x2
+ x3 +x4
x1 + x2 + x3 + x4
2
2
f
=f
≤
4
2 14 FELIX LAZEBNIK
+ f x3 +x4
2
≤
2
f (x1 )+f (x2 )
+ f (x3 )+f (x4 )
2
2
=
2
f (x1 ) + f (x2 ) + f (x3 ) + f (x4 )
.
4
Hence, the statement is proven for n = 22 . Similarly, for n = 23 = 8, we have
x1 +...+x4
+ x5 +···+x8
x1 + . . . + x8
4
4
f
=f
≤
8
2
f x1 +...+x4 + f x5 +...+x8
4
4
≤
2
f (x1 )+...+f (x4 )
+ f (x5 )+...+f (x8 )
4
4
=
2
f (x1 ) + . . . + f (x8 )
.
8
The transition from n = 2k to n = 2k+1 follows similarly to the preceding particular
cases. As
x k +...+x k+1
x1 +...+x2k
+ 2 +1 2k 2
x1 + . . . + x2k+1
2k
=
2k+1
2
for all k ≥ 2, we have:
x1 +...+x
x k +...+x k+1
2k
+ 2 +1 2k 2
x1 + . . . + x2k+1
2k
f
=f
≤
2k+1
2
x
+...+x k+1
x +...+x
k
f 1 2k 2k + f 2 +1 2k 2
≤
2
f x 1 +x2 2 f (x1 )+...+f (x2k )
2k + f (x2k +1 )+...+f (x2k+1 )
2k =
2
f (x1 ) + . . . + f (x2k+1 )
.
2k+1
Now we suppose that the statement is proven for n ≥ 3 values of xi ’s and show
that it implies the statement for n − 1 values of x (the ‘down’ part). Let xi ∈ (a, b),
i ∈ [n − 1], be arbitrary n − 1 numbers on (a, b). Apply the inequality to the
following n numbers:
n−1
xi
x1 + . . . + xn−1
x1 , x2 , . . . , xn−1 , and xn =
= i=1 .
n−1
n−1
We have
n
n
f (xi )
i=1 xi
f
≤ i=1
⇔
n
n
n−1
n n−1 n−1
i=1 xi
xi
i=1 f (xi ) + f
n−1
i=1
n−1
f
≤
⇔
n
n
n−1
n−1
i=1 xi
n−1
f (xi ) + f
i=1
n−1
i=1 xi
f
≤
.
n−1
n NOTES ON INEQUALITIES 15 n−1
i=1 xi
Solving the last inequality for f
, we obtain
n−1
n−1
n−1
f (xi )
i=1 xi
f
≤ i=1
.
n−1
n−1
This completes the proof. The assertion about the equality sign should be a part
of the inductive hypothesis, and it follows immediately.
Several famous inequalities can be obtained as simple corollaries of Jensen’s
inequality:
• f (x) = x2 gives us that the quadratic mean is greater or equal than the
arithmetic mean:
n i=1 xi ≤ n i=1 x2
i 1 / 2 .
n
n
The quadratic mean of numbers x1 , x2 , . . . , xn is the name of the number
on the right. This inequality is called the arithmeticquadratic mean
inequality. This inequality is useful in Probability theory and Statistics.
• f (x) = ln x, x > 0 gives that the arithmetic mean of n positive real numbers
is greater or equal than the geometric mean:
n i=1 xi n ≥ n
xi i=1 1/n . This inequality is called the arithmeticgeometric mean inequality
(AGM). It allows to minimize the sum of numbers whose product is ﬁxed,
or to maximize the product of numbers whose sum is ﬁxed.
• f (x) = 1/x, x > 0 gives
n i=1 n xi ≥ n n 1
i=1 xi . The expression on the right is called the harmonic mean of n positive
numbers, and the inequality is called the arithmeticharmonic mean
inequality. An equivalent way to write it is:
n
n
1
xi
≥ n2 .
xi
i=1
i=1
In order to apply Jensen’s inequality, one has to check that the function is concave
up (down). This can be done by using the deﬁnition (n = 2), or by using the second
derivative test when it is applicable. Let’s do it for the three function above.
1. f (x) = x2 . Using the deﬁnition, we have:
x + x x + x 2
x2 + x2
f (x1 ) + f (x2 )
1
2
1
2
2
f
=
≤1
=
,
2
2
2
2 16 FELIX LAZEBNIK and the equality is attained only if it is attained in
x + x 2
x2 + x2
1
2
2
≤1
⇔ (x1 − x2 )2 ≥ 0,
2
2
i.e., for x1 = x2 .
Using the second derivative, we have: f (x) = 2 > 0 for all real x. Each
argument proves that f is convex up on R.
2. f (x) = ln x, x > 0. Using the deﬁnition, we have:
x + x
x + x
1
2
1
2
f
= ln
,
2
2
√
f (x1 ) + f (x2 )
ln x1 + ln x2
1
=
= ln(x1 x2 ) = ln( x1 x2 ),
2
2
2
x + x
√
√
√
1
2
ln
≥ ln( x1 x2 ) ⇔ ( x1 − x2 )2 ≥ 0.
2
Hence,
x + x f (x ) + f (x )
1
2
1
2
f
≥
,
2
2
and the function is convex down. The equality is attained only if x1 = x2 > 0.
Using the second derivative, we have: (ln x) = (1/x) = −1/x2 < 0 for all
x > 0. Hence, f is convex down on (0, ∞). 3. f (x) = 1/x, x > 0. We leave it to the reader to show that f is convex up on
(0, ∞).
6. The ArithmeticGeometric Mean Inequality (AGM)
Here we wish to present several more examples of applications of the AGM
inequality: For all nonnegative real numbers x1 , x2 , . . . , xn ,
√
x1 + x2 + . . . + xn
n
x1 x2 · · · xn ≤
,
n
with equality taking place if and only if all xi are equal.
Example 11. What is the maximum product of four positive numbers which add
to 20? Which numbers maximize the product?
Solution. Let a, b, c, d ≥ 0 be the numbers. Then a + b + c + d = 20. By the
AGM inequality,
√
a+b+c+d
4
abcd ≤
= 5.
4
Hence abcd ≤ 54 = 625, with equality if and only if a = b = c = d (= 5). Hence,
the greatest value of the product is 625, and it is attained if and only if a = b =
c = d = 5.
Those who are familiar with calculus of several variables may try to apply it to
the Example 11. It will not work as fast. NOTES ON INEQUALITIES 17 Example 12. Out of all rectangular parallelepipeds (boxes) of unit volume, which
one has the smallest surface area?
Solution. Let x, y, z > 0 denote the dimensions of the parallelepiped. Then
its volume is xyz = 1, and its surface area is A = 2(xy + yz + zx). As numbers
xy, yz, zx are positive, the AGM inequality gives:
√
xy + yz + zx
≥ 3 xy · yz · zx = 3 (xyz )2 = 1.
3
Hence, xy + yz + zx ≥ 3, and A ≥ 6. The equality is achieved if and only if xy =
yz = zx. As x, y, z > 0, we obtain x = y = z (= 1). Therefore he parallelepiped of
minimum surface area is the unit cube.
Example 13. Show that 2x3 + 5x + 1
x4 > 6 for all positive x. Solution. Applying the AGM inequality for three positive numbers 2x3 , 5x, and
we obtain:
1
√
2x3 + 5x + x4
1
3
3
.
10 = 2x3 · 5x · 4 ≤
x
3
√
√
3
1
Therefore 3 10 ≤ 2x3 + 5x + x4 . Now observe that 6 < 3 3 10 as 63 = 216 <
√3
(3 3 10) = 270.
1
x4 , Is it easy to obtain the result of Example 13 by using Calculus?
n
1
Example 14. Prove that for all integer n ≥ 2, 1 + n
< 1+ 1
n+1 n+1 . Proof. Consider n + 1 positive numbers: x1 = x2 = . . . = xn = 1 + 1/n, and
xn+1 = 1. Applying the AGM inequality to these numbers, we obtain
1
1
n+1
n · (1 + n ) + 1
1
(1 + )n · 1
<
n
n+1
n + 2 n+1
1 n
<
1+
n
n+1
1 n
1 n+1
1+
<
1+
.
n
n+1 18 FELIX LAZEBNIK 7. Problems All letters in these exercises represent REAL numbers, and we will not be repeating this. Many of the problems below can be solved without references to Calculus,
but you are welcome to use Calculus as much as you wish. Problems which require
c
use of Calculus (or greatly beneﬁt from it) are marked by red symbol . Most
problems can be approached by the methods discussed in this paper, such as using
the elementary properties of inequalities, case analysis, using the the method of
intervals, the method of mathematical induction, Jensen’s inequality and its corollaries (especially AGM or AQM inequalities). Horizontal lines separate problems
of various diﬃculties (in the opinion of the author): easier, medium, harder.
1. Prove each correct statement. Give a counterexample to each false statement. If you see a short way to modify false statements such that they
become correct ones, please do it.
(a) For all a, b, c, d, a > b and c > d imply a − c > b − d.
(b) For all a, b, c, d, a > b and c > d imply ac > bd.
(c) For all a, a2 > a3 .
(d) For all a > 1, 1/a < 1.
(e) For all nonzero a, b, if a > b then 1/a < 1/b.
(f) For all positive a, b, c, d, a > b and c > d implies a/c > b/d.
(g) For all a, b, a > b implies a2 > b2
(h) For all a, b, a2 > b2 implies a > b.
(i) For all a, b, a3 > b3 if and only if a > b.
(j) For all positive integer n, and all a, b, a2n+1 > b2n+1 if and only if
a > b.
√
(k) For all a ≥ 0 and all b, a √ b implies a > b2 .
>
(l) For all a, b, a > b2 implies a > b.
√
(m) For all a ≥ 0 and all b, a < b implies√ < b2 .
a
(n) For all a ≥ 0 and all b, a < b2 implies a < b.
2. Justify the following statements.
(a) If x > 1, then x5 > x2 .
(b) If 0 < x < 1, then x10 < x2 < x10 + 1.
(c) If f (x) = x2 − 5x − 1, then f (1234567) > 0, and f (0.1234567) < 0.
3. Without using calculator, or computers, decide which number is greater
and prove your answer. (But you are welcome to use anything you wish in
order to check whether your answer is correct.)
√
√
√
√
(a) 2 2 or 3 3 ? 4 4 or 5 5
√
√
√
√
(b) 10 + 30 or 11 + 29
√ (c) 1/4 or √
5 (d) √
3 10 + √ 5+1
√
10−2 5 8 or 5 NOTES ON INEQUALITIES 19 4. (i) Without using a calculator or computer, explain which of the two numbers is larger:
21,000,000 or 1, 000, 0001,000 .
c
(ii) Do graphs of the curves y = 2x and y = x1000 intersect over
[3, ∞)? 5. A boat travels in the river from A to B and then back to A. The speed of
a boat in still water is v miles per hour, and the speed of the current is u
miles per hour (u < v ). Suppose t1 is the time of the round trip between A
and B if there were no current, and t2 is the time of the round trip between
A and B with the current. Compare t1 and t2 .
6. Solve the following inequalities.
(a) (i) x2 − 3x − 5 ≥ 0 (b) (i) 2 x +3x−10
x2 −3x−28 <0 (c) (i) 2x − 4 ≥ 5 (ii) x2 − 3x + 5 ≥ 0 (ii) 2 x −25
(x−2)(x2 −x+6) (ii) 3x − 4 + x ≤ 6 ≥0 (iii) 2x2 − x + 1 < 0
(iii) x2 − 4x ≥ 6 (d) (i) x + 5 + x − 11 > 20
(ii) x + 5 + x − 11 < 15
x+2
x−1
x+1
> 3.
(e) (i) x+2 > x+5 .
(ii) x−1
(f) 5 − 3x < (x − 1)(2 − x)
√
(g) 2 − x ≤ x + 1. 7. Use the method of mathematical induction to show that for n ≥ 2, inequalities a1 < b1 , a1 < b1 , . . ., an < bn with all ai positive, imply
a1 a2 . . . an < b1 b2 . . . bn . Conclude from here that 0 < a < b implies an < bn
for all integer n ≥ 2.
8. Prove the following inequalities. (a) 2x3 > x + 1 if x > 1, and 2x3 < x + 1 if x < 1.
(b) For all x, x12 − x9 + x4 − x + 1 > 0.
(c) For all x, 2x4 + 1 ≥ 2x3 + x2 . 9. Let a, b be the lengths of two legs of a right triangle, and let c be the length
of its hypothenuse. Prove that a3 + b3 < c3 .
10. Find all solutions of the equation cos2008 x + sin2008 x = 1 on [0, 2π ].
11. Let x1 , , x2 , . . . , xn be real numbers, let m be the smallest of them, and M
be the largest of them. Prove that
x1 + x2 + . . . + xn
m≤
≤ M,
n
with equality in each part if and only if all n numbers are equal.
12. Prove the following inequalities.
b
(a) If ab > 0, then a + a ≥ 2. Describe completely when the equality
b
takes place.
b
c
(b) If a, b, c > 0, then a + c + a ≥ 3, and the equality occurs if and only
b
if a = b = c.
(c) For all x, y, z , x2 + y 2 + z 2 ≥ xy + yz + zx, and the equality occurs if
and only if x = y = z . 20 FELIX LAZEBNIK (d) If x + y + z = 1, then x2 + y 2 + z 2 ≥ 1/3. Describe completely when
the equality takes place.
(e) If not all a, b, c are equal, then a2 + b2 + c2 + 3 > 2(a + b + c).
(f) Suppose a1 a2 . . . an = 1 and all ai > 0. Prove that (1 + a1 )(1 +
a2 ) . . . (1 + an ) ≥ 2n . Describe completely when the equality takes
place.
13. Out of all rectangular parallelepipeds
(i) of the given perimeter, which one has the largest volume?
(ii) of the given volume, which one has the smallest perimeter?
(iii) of the given surface area, which one has the largest volume?
14. Let a, b, c be the length of sides of a triangle ABC , and let p = (a + b + c)/2
be its semiperimeter. Then the area of the triangle can be found by the
famous ArchimedesHeron formula:
Area ABC = p(p − a)(p − b)(p − c). Use the AGM inequality to prove that
(i) out of all triangles with a given perimeter (= 2p), the equilateral
triangle has the greatest possible area.
(ii) out of all triangles with a given area, the equilateral triangle has the
smallest perimeter. 15. Prove that for n ≥ 5, 2n > n2 16. Prove that for n ≥ 0, 3n > (n − 1)3 17. Investigate for which positive integers n, n! > n2n , and prove your result.
18. Investigate for which positive integers n, 3n > 10n2 , and then prove your
result
19. Prove that for n ≥ 2, (1 + 1/3)n > 1 + n/3 20. (Bernulli’s Inequality) Prove that for any ﬁxed real number x, −1 < x = 0,
and every integer n ≥ 2,
(1 + x)n > 1 + nx. This inequality is useful to provide a simple rough lower bound on the values
of exponential functions. For example, what does it imply about the values of
1.0002100 or .99951000 ? 21. Let x1 , x2 , . . . , xn be n real numbers, n ≥ 2. Prove that x1 + x2 + · · · + xn  ≤ x1  + x2  + · · · + xn . 22. Let x1 , x2 , . . . , xn be n real numbers, n ≥ 2. Prove that  sin(x1 + x2 + · · · + xn ) ≤  sin x1  +  sin x2  + · · · +  sin xn . 23. Determine whether the following functions are convex up, or down, or neither on the given intervals. In all parts, using the second derivative test
provides a faster solution.
(a) f (x) = cos x on [0, π /2].
(b) f (x) = cos x on [0, π ].
c
(c) f (x) = tan x on [0, π /2).
x
(d) f (x) = 1−x on (1, ∞). NOTES ON INEQUALITIES 21 100
c
(e) f (x) = k=1 n xk on (−1, ∞).
k
c
(f) f (x) = x ln x on (0, ∞). 24. Verify that the function f (x) = x3 is convex up on (0, ∞). Apply Jensen’s
inequality to function f and its values at three positive real numbers a, b, c.
What inequality do you obtain?
25. Verify that the function f (x) = sin x is convex up on [0, π ]. Apply Jensen’s
inequality to function f and its values at α, β , γ from [0, π ] . What inequality do you obtain?
26. Let f : (a, b) → R and g : (a, b) → R be two functions convex up on (a, b).
(i) Is f + g be convex up on (a, b)?
c
(ii) Is f · g convex up on (a, b)?
27. Let f : (a, b) → (0, ∞) be a convex up function on (a, b).
(i) Is 1/f convex up on (a, b)?
(ii) Is 1/f convex down on (a, b)?
28. Prove that for n ≥ 2, 1
1
1
+
+ ... +
> 1/2.
n+1 n+2
2n This inequality leads to an easy proof of the famous fact that the sum 1/1 +
1/2 + 1/3 + . . . + 1/n can exceed any ﬁxed number provided n being suﬃciently
large. 29. Prove that for every positive integer n,
1
1
1
1
1
+ 2 + 2 + ··· + 2 ≤ 2 − .
12
2
3
n
n
This inequality implies that the sum 1/12 + 1/22 + 1/32 + · · · + 1/n2 cannot
exceed 2 no matter how large n is. It was proven by L. Euler that as n becomes
larger, this sum becomes closer and closer to π 2 /6. 30. Prove that for every positive integer n,
√
1
1
1
1
√ + √ + √ + · · · + √ > 2 n + 1 − 2.
n
1
2
3
31. Prove that for n ≥ 2, 1
135
2n − 1
1
√ < · · · ... ·
<√
·
246
2n
2n
3n + 1 32. Prove that for every two nonnegative real numbers a and b, and every
integer n ≥ 2,
n
a+b
an + bn
≤
,
2
2
where the equality is attained if and only if a = b. 22 FELIX LAZEBNIK 33. Use the AGM inequality. Let the sum of positive numbers x, y, z be 60.
Use the AGM inequality to determine the maximum value of
(i) (x − 3)(y + 1)(z + 5)
(ii) (x − 3)(2y + 1)(3z + 5).
√
34. Consider a sequence {an }n≥1 , where a1 = 2 and an+1 = 2an + 5 for
n ≥ 1.
(i) Prove that an+1 > an for all n ≥ 1.
(ii) Prove that an < 4 for all n ≥ 1.
√
35. Consider a sequence {bn }n≥1 , where b1 = 2 and bn+1 = bn + 2 for n ≥ 1.
(i) Prove that bn+1 < bn for all n ≥ 1.
(ii) Prove that bn > 2 for all n ≥ 1.
36. Prove that
(i) sin t < t for all t > 0;
(ii) tan t > t for t ∈ (0, π /2). c
37. (i) Let f : [a, ∞) → R be a continuous function such that f (a) ≥ 0 and
f (x) > 0 for all x > 0. Prove that f (x) > 0 for al x > 0.
(ii) Let f, g : [a, ∞) → R be continuous functions such that f (a) ≥ g (a)
and f (x) > g (x) for all x > 0. Prove that f (x) > g (x) for al x > 0.
c
38. Prove that
(a) cos x > 1 − x2 /2 for all x ∈ R.
(b) sin x > x − x3 /6, for all x > 0.
(c) cos x ≥ 1 − 2x on [0, π /2].
π
(d) ex > 1 + x + x2 for all x ∈ R.
(e) x − x2 /2 < ln(1 + x) < x for all x > 0.
c
39. Prove that for for all 0 < x < y < π /2,
y
x
(a) sin x < sin y , x − tan x > y − tan y . c
40. The Mean Value Theorem states that for any function continuous on [a, b]
and diﬀerentiable on (a, b), there exists c ∈ (a, b), such that f (b) − f (a) =
f (c)(b − a). Use this theorem to show that
(i)  sin x − sin y  ≤ x − y  for any x, y ∈ R.
(ii)  tan x − tan y  ≤ x − y  for any x, y ∈ (0, π /2).
41. What is the greatest value of sin A + sin B + sin C , where A, B, C are measures of three angles of a ABC ?
c
42. Use Jensen’s inequality to prove that for any a, b, c on (0, ∞),
a+b+c
a+b+c
aa bb cc ≤
.
3 43. Use Jensen’s inequality to prove that for any n ≥ 2 real numbers x1 , x2 , . . . , xn
on (1, ∞),
n
n
xi
i=1 x
n i ≥
n·
.
1 − xi
n − i=1 xi
i=1 NOTES ON INEQUALITIES 23 c
44. Show that if f (x) > 0 (f (x) < 0) for all x ∈ (a, b), then f is convex
up (down) on (a, b).
c
45. What is greater:
√
√
(i) 78 78 or 79 79? (ii) eπ or π e ? 1
c
46. The goal of this problem is to show that the function y = f (x) = (1+ x )x
is increasing on I = [1, ∞). This result can be used to justify the fact that
when the interest in the bank is compounded more times per year, the
better this is for a customer. (a) Find y (x). Is it obvious that y (x) > 0 for all x in I ?
2
(b) Prove that ln(1 + t) > t − t2 for any t > 0.
1
1
1
(c) Prove that x − 2x2 > x+1 for x > 1.
1
1
(d) Prove that ln(1 + x ) > x+1 for x > 1.
(e) Conclude that f is increasing on I .
Compare this problem with Example 14, where a similar statement had
to be proved for values of the sequence f (n), n = 1, 2, 3, . . ..
47. Let a1 < a2 < . . . < an be ﬁxed numbers, and
f (x) = x − a1  + x − a2  + . . . + x − an . Find minx∈R f (the minimum value of f over all reals), and all value(s) of
x for which this minimum is attained. What about maxx∈R f ? 48. Let a, b, c be positive real numbers such that abc = 1. Prove that
a(a − 1) + b(b − 1) + c(c − 1) ≥ 0. 49. Prove that out of all ngons, n ≥ 3, inscribed in a circle, the regular one
has the greatest area. You can assume without proof that the center of the
circle is inside the ngon.
50. Prove that for all n ≥ 2, 1
2 < (1 + n )n < 3. 51. Represent 1000 as a sum of several positive integers which product is the
greatest. (The number of integers is not given.) 24 FELIX LAZEBNIK 8. Hints and Answers to Problems These are NOT complete solutions. Just some answers and hints.
1. In order to show that the statement “for all a, b, c, d, . . . A → B ” is false,
one can show that there exists some values of a, b, c, d, . . . that A is true
and B is false.
(a) False.
(b) False. True if all parts of all inequalities are positive.
(c) False.True if . . . 0 = a < 1
(d) True.
(e) False.
(f) False. A counterexample:
(g) False. True if . . .
(h) False. True if . . .
(i) True.
(j) True.
(k) False: take a = 1 and b = −2. True if . . .
(l) True.
(m) True.
(n) True.
2. (a) Hint: x5 − x2 = x2 (x3 − 1).
(b) Hint: consider each part of the inequality separately.
(c) Hint: use idea of part (a).
3. (a) Hint: Raise both numbers we compare to the nth power for some
special values of n.
(b) Hint: Begin by squaring both numbers.
(c) Hint: Begin by squaring both numbers.
√
(d) Hint: First subtract 8 from both numbers.
4. (i) Hint: use laws of exponents.
(ii) Answer: yes.
5. Answer: t1 < t2 .
√ √ 6. (a) Answers: (i) (−∞, 3−2 29 ) ∪ ( 3+2 29 , ∞) ; (ii) R; (iii) ∅.
(b) Answers: (i) (−5, −4) ∪ (2, 7); (ii) [−5, 2) ∪ [5, ∞).
(c) Answers: (i) (−∞, −0.5] ∪ [4.5, ∞); (ii) [−1, 2.5];
√
√
(iii) (−∞, 2 − 10] ∪ [2 + 10, ∞).
(d) Answers: (i) (−∞, −7) ∪ (13, ∞); (ii) ∅.
(e) Answers: (i) (−5, −2) ∪ (7, ∞); (ii) (0.25, 1) ∪ (1, 2.5).
(f) Answer: (1.5, 2] √
(g) Answer: ((−3 + 13)/2, 2] 7. Straightforward. 8. (a) Hint: 2x3 = x3 + x3
(b) hint: Consider three cases: x ≤ 0, 0 < x < 1, and x ≥ 1. NOTES ON INEQUALITIES 25 (c) Factor the polynomial f (x) = 2x4 − 2x3 − x2 + 1 ≥ 0. Or use Calculus
to ﬁnd minx∈R f (x). 9. Hint: use the Pythagoras theorem in the form (a/c)2 + (b/c)2 = 1.
10. Answer: {0, π /2, π , 3π /2, 2π }. 11. Hint: Replace each number with m to get the ﬁrst inequality.
12. (a) Very simple.
(b) Use the AGM inequality.
(c) Hint: complete the square with respect to each variable, or apply the
obvious inequality (a − b)2 ≥ 0 three times. Another approach is to
“symmetries”: write x = 1/3 + α, y = 1/3 + β , z = 1/3 + γ . The
third approach is to ﬁnd the distance from the origin to the plane
x + y + x = 1.
(d) Hint: several diﬀerent ideas will work. For example, one can use the
inequality from part (c).
(e) Easy.
√
(f) Hint: ai + 1 ≥ 2 ai . The equality takes place if and only if ai = 1.
13. Use the AGM inequality.
14. Use the AGM inequality.
15. Use the Method of Mathematical Induction.
16. Use the Method of Mathematical Induction. The usual logic of showing
A > C given A > B and proving B ≥ C , will work here beginning with
n ≥ 4, though the inequality holds for n = 0, 1, 2, 3 as well. This illustrates
the suﬃciency of showing B ≥ C , rather than its necessity.
17. Substitute several ﬁrst values of n: n = 1, 2, . . . . Use the Method of Mathematical Induction.
18. Substitute several ﬁrst values of n: n = 1, 2, . . . . Use the Method of Mathematical Induction.
19. Use the Binomial formula, or use the Method of Mathematical Induction.
20. Use the Method of Mathematical Induction.
21. Use the Method of Mathematical Induction. The most important is the
base case: n = 2.
22. Use the formula for the sine of the sum of two angles, and the the Method
of Mathematical Induction. The most important is the base case: n = 2.
23. (a)
(b)
(c)
(d)
(e)
(f) Answer:
Answer:
Answer:
Answer:
Answer:
Answer: convex down. Use trigonometric formuli.
not convex.
convex up. Use Calculus.
convex down.
convex up. Use the Binomial Theorem.
convex up. Use Calculus. 24. Hint: Easy.
25. Hint: similar to Problem 23 (a). 26 FELIX LAZEBNIK 26. (i) Answer: yes. (ii) Answer: no. Find a counterexample.
27. Let f : (a, b) → (0, ∞) be a convex up function on (a, b). (i) Answer: no.
Find a counterexample. (ii) Answer: no. Find a counterexample.
28. Compare each term with 1/2. Or use the Method of Mathematical Induction.
1
1
1
1
29. Hint: Check that k2 < (k−1)k = k−1 − k , and use this fact. Or use the
Method of Mathematical Induction. 30. Use the Method of Mathematical Induction.
31. Use the Method of Mathematical Induction.
32. Use the Method of Mathematical Induction. Consider two cases: a ≤ b
and a > b. Another approach is to use the second derivative test for the
function y = xn on [0, ∞). 33. Use the AGM inequality. Hints: (i) Use the AGM inequality. (ii) 2y + 1 =
2(y + 1/2), 3z + 5 = 3(z + 5/3).
34. Hints: (i), (ii) Use the Method of Mathematical Induction.
35. Hints: (i), (ii) Use the Method of Mathematical Induction.
36. (i) Let 0 < t ≤ 1, and let Pt : (cos t, sin t) be the point on the unit circle
x2 + y 2 = 1 corresponding to the angle t radians. Let O be the origin.
Compare the area of OPt P0 with the area of the sector OPt P0 . For t > 1,
the statement is obvious.
(ii) Let 0 < t < π /2, and let Pt : (cos t, sin t) be the point on the unit circle
x2 + y 2 = 1 corresponding to the angle t radians. Let O be the origin,
and let M be the point of intersection of the line OPt with the line x = 1.
Compare the area of the sector OPt P0 with the area of the triangle OM P0 .
37. (i) Hint: Use the Mean Value Theorem.
(ii) Hint: let h(x) = f (x) − g (x). Apply part (i) to the function h. 38. Hint: use the statement from Problem 37.
39. No other hints.
40. No other hints.
41. Use Jensen’s inequality. 42. Hint: consider the function y = x ln x on (0, ∞).
43. Hint: consider the function y = x
1−x on (0, ∞). 44. Hint: a proof can be found in most Calculus books.
45. Hint: Consider the logarithms of the numbers. Use Calculus.
46. Use the statement from Problem 37.
47. Hint: Think about a − b geometrically. Experiment with n = 2, 3, 4. NOTES ON INEQUALITIES 27 48. Use AGM and Jensen’s inequalities.
49. Hint: Use Jensen’s inequality.
50. Hint: The ﬁrst inequality is easy. For the second one, use the Binomial
formula, and bound from above the sum of all terms starting from the
third by 1/2 + 1/22 + . . . + 1/2n . Then show that this sum is less than 1.
51. Hint: ﬁrst experiment with a similar question, when 1000 is replaced by 9,
or 10, or 11, or 12. Use the fact that for an integer a ≥ 5, 2 · (a − 2) > a,
and that 32 > 23 .
•••••••
These notes represent a very brief introduction to the vast subject of inequalities.
Much more can be found in the references below. References
[1] E. Beckenbach, R. Bellman, An Introduction to Inequalities, Random House, 1961.
(A great introduction to inequalities. Covers more than these notes. Can be easily understood
by good high school students. No use of Calculus.)
[2] E. Beckenbach, R. Bellman, Inequalities, SpringerVerlag, 1961.
(An advanced version of [1] for college students. Begins with twelve proofs of the AGM.)
[3] G. H. Hardy, J. E. Littlewood, and G. Plya, Inequalities, Cambridge University Press; 2
edition 1988.
(A classic. For advanced college students.)
[4] K. Kedlaya, A < B , http://www.artofproblemsolving.com/Resources/Papers/KedlayaInequalities.pdf.
A nice collection of various examples and techniques. More advanced than these notes. Based
on notes for Mathe Olympiad Program (MOP), 1999.
[5] V.G. Kovalenko, M.B. Gel’fand, R.P. Ushakov, Proofs of Inequalities, Vyscha Shkola, 1979.
(In Ukrainian).
(The most elementary introduction to the subject. Covers several other famous inequalities
and related techniques from Calculus.)
[6] D. O. Shklarsky, N. N. Chentzov, and I. M. Yaglom , Geometric Inquality and Problems on
Maximum and Minimum, Nauka, 1970. (In Russian).
(A great discussion of inequalities related to geometry. Can be easily understood by good high
school students. No use of Calculus.)
[7] I.H. Sivashinskiy, Inequalities through Problems, Nauka, Moscow, 1967. (In Russian).
(A very readable discussion of various questions about inequalities. Can be easily understood
by good high school students. No use of Calculus.)
[8] V.M. Tikhomirov, Stories about maxima and minima, MAA, 1990.
(An outstanding collection of ﬁfteen “stories” related to maxima and minima problems. A
very useful reading for everyone who likes mathematics and who knows Calculus and a little
beyond it. ) ...
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This note was uploaded on 12/07/2011 for the course MATH 245 taught by Professor Cioaba during the Fall '10 term at University of Delaware.
 Fall '10
 CIOABA
 Calculus, Inequalities

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