HOMEWORK 2: COMMENTS
1.
Nonbook problems
(2)
(a) Since
p

a
and
p

b
there exist integers
q
1
and
q
2
such that
a
=
pq
1
and
b
=
pq
2
.
Hence
c
=
p
(
q
1

q
2
)
,
and so
p

c
.
(b) It does not follow.
For example, let
a
= 13
b
= 1
, c
= 12 and
p
= 2.
Then
a
=
b
+
c
and
p

c
, yet
p
does not divide
a
.
(4)
(b) We may apply the result of part (a) three times to conclude that
x
2
+
y
2
≥
2
xy,
y
2
+
z
2
≥
2
yz,
and
z
2
+
x
2
≥
2
zx.
Adding these three inequalities we obtain
2(
x
2
+
y
2
+
z
2
)
≥
2(
xy
+
yz
+
zx
)
.
The result now follows after we divide the above inequality by 2.
(5) Recall we proved the following lemma in class:
Lemma 1.1.
Let
x, y, z
and
w
be nonnegative real numbers. Suppose that
x
≥
z
and
y
≥
w
. Then
xy
≥
zw
.
Since
a, b
and
c
are nonnegative real numbers, it follows from the AGmean in
equality that
(
a
+
b
)
≥
2
√
ab,
(
b
+
c
)
≥
2
√
bc,
and
(
c
+
a
)
≥
2
√
ca.
We may multiply the above inequalities (thanks to Lemma 1.1, applied twice) to
obtain
(
a
+
b
)(
b
+
c
)(
c
+
a
)
≥
8
√
ab
√
bc
√
ca
= 8
abc.
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 Spring '07
 COURTNEY
 Addition, Division, Integers, Negative and nonnegative numbers, positive integers

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