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Here is the details: First dividing 4 on both sides and moving 27(4 +
β
)
2
on the right hand side:
4(6 +
β
)
3
27(4 +
β
)
2
≥
4
⇐⇒
(6 +
β
)
3
≥
27(4 +
β
)
2
(1)
Then we may do a sub
u
= 4 +
β
, then we get
(6 +
β
)
3
≥
27(4 +
β
)
2
⇐⇒
(
u
+ 2)
3
≥
27
u
2
(2)
then expand (
u
+ 2)
3
=
u
3
+ 6
u
2
+ 12
u
+ 8, then we get
(
u
+ 2)
3
≥
27
u
2
⇐⇒
u
3
+ 6
u
2
+ 12
u
+ 8
≥
27
u
2
(3)
Then move 27
u
2
on the left hand side
u
3
+ 3
u
2
+ 12
u
+ 8
≥
27
u
2
⇐⇒
u
3

21
u
2
+ 12
u
+ 8
≥
0
(4)
Then our goal is to solve the cubic inequality(we may just take =, our goal is to ﬁnd the smallest
β
), there is no formula for us to solve it directly. So we usually guess a solution for this, it turns out
that
u
= 1 is a solution, that means we ﬁnd one factor ((
u

1)) for the left hand side of (5)
u
3

21
u
2
+ 12
u
+ 8 = 0
(5)
Here we may write down
u
3

21
u
2
+ 12
u
+ 8 = (
u

1)(
au
2
+
bu
+
c
)
(6)
here
a,b,c
are constants. You may expand right hand side of (7) and then compare the coef. of both
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This note was uploaded on 01/11/2012 for the course MATH 334 taught by Professor Dallon during the Fall '08 term at BYU.
 Fall '08
 DALLON
 Differential Equations, Equations

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