can yeu hang
DESCRIPTION
InequalityTRANSCRIPT
1. Let a, b, c be nonnegative real numbers, now two of which are zero. Provethat
a3
a2 + b2+
b3
b2 + c2+
c3
c2 + a2≥√
3(a2 + b2 + c2)2
.
Solution. Rewrite the inequality as∑cyc
(a3 + b3
a2 + b2− a + b
2
)≥√
3∑cyc
a2 −∑cyc
a +∑cyc
b3 − a3
a2 + b2
⇔∑cyc
(a− b)2(a + b)2(a2 + b2)
≥∑cyc
(a− b)2√3∑cyc
a2 +∑cyc
a
+
(a− b)(b− c)(c− a)
(∑cyc
a2b2 + abc∑cyc
a
)(a2 + b2)(b2 + c2)(c2 + a2)
Since√
3∑cyc
a2 ≥∑cyc
a, it suffices to prove
∑cyc
(a− b)2(a + b)2(a2 + b2)
≥∑cyc
(a− b)2
2∑cyc
a+
(a− b)(b− c)(c− a)
(∑cyc
a2b2 + abc∑cyc
a
)(a2 + b2)(b2 + c2)(c2 + a2)
⇔∑cyc
(a−b)2(
a + b
a2 + b2− 1
a + b + c
)≥
2(a− b)(b− c)(c− a)
(∑cyc
a2b2 + abc∑cyc
a
)(a2 + b2)(b2 + c2)(c2 + a2)
⇔∑cyc
(a−b)2·2ab + ac + bc
a2 + b2≥
2(a− b)(b− c)(c− a)
(∑cyc
a
)(∑cyc
a2b2 + abc∑cyc
a
)(a2 + b2)(b2 + c2)(c2 + a2)
By the AM-GM Inequality, we have∑cyc
(a− b)2 · 2ab + ac + bc
a2 + b2
≥ 3 3
√(a− b)2(b− c)2(c− a)2(2ab + ac + bc)(2bc + ab + ac)(2ac + bc + ba)
(a2 + b2)(b2 + c2)(c2 + a2)
1
It remains to prove that
3 3
√(a− b)2(b− c)2(c− a)2(2ab + ac + bc)(2bc + ab + ac)(2ac + bc + ba)
(a2 + b2)(b2 + c2)(c2 + a2)
≥2(a− b)(b− c)(c− a)
(∑cyc
a
)(∑cyc
a2b2 + abc∑cyc
a
)(a2 + b2)(b2 + c2)(c2 + a2)
⇔ 27
[∏cyc
(2ab + ac + bc)
][∏cyc
(a2 + b2)2]
≥ 8
[∏cyc
(a− b)
](∑cyc
a
)3(∑cyc
a2b2 + abc∑cyc
a
)3
Since ∏cyc
(2ab + ac + bc) ≥ 2
(∑cyc
ab
)3
and ∏cyc
(a2 + b2)2 ≥ 6481
(∑cyc
a2
)2(∑cyc
a2b2
)2
It suffices to show
163
(∑cyc
ab
)3(∑cyc
a2
)2(∑cyc
a2b2
)2
≥
[∏cyc
(a− b)
](∑cyc
a
)3(∑cyc
a2b2 + abc∑cyc
a
)3
Now, note that
8
(∑cyc
a2b2
)2(∑cyc
ab
)2
− 3
(∑cyc
a2b2 + abc∑cyc
a
)3
= 8
(∑cyc
a2b2
)2(∑cyc
a2b2 + 2abc∑cyc
a
)− 3
(∑cyc
a2b2 + abc∑cyc
a
)3
= A
(∑cyc
a2b2 − abc∑cyc
a
)≥ 0
where
A = 5
(∑cyc
a2b2
)2
+ 12abc
(∑cyc
a2b2
)(∑cyc
a
)+ 3a2b2c2
(∑cyc
a
)2
2
It remains to prove that
2
(∑cyc
ab
)(∑cyc
a2
)2
≥
[∏cyc
(a− b)
](∑cyc
a
)3
Without loss of generality, we may assume a + b + c = 1. Setting q = ab + bc +ca, r = abc, then
(a−b)(b−c)(c−a) ≤√
(a− b)2(b− c)2(c− a)2 =√
q2 − 4q3 + 2(9q − 2)r − 27r2
we have to prove
2q(1− 2q)2 ≥√
q2 − 4q3 + 2(9q − 2)r − 27r2
If 9q ≤ 2, then
2q(1− 2q)2 −√
q2 − 4q3 + 2(9q − 2)r − 27r2 ≥ q[2(1− 2q)2 −
√1− 4q
]≥ 0
Since
2(1− 2q)2 −√
1− 4q =(√
1− 4q − 12
)2
+14
[2(1− 4q)2 + 1] ≥ 0
If 9q ≥ 2, then
√q2 − 4q3 + 2(9q − 2)r − 27r2 =
√427
(1− 3q)3 − 127
(27r − 9q + 2)2
≤√
427
(1− 3q)3
⇒2q(1− 2q)2 −√
q2 − 4q3 + 2(9q − 2)r − 27r2
≥ 2q(1− 2q)2 −√
427
(1− 3q)3 = 2q(1− 2q)2 − 29
(1− 3q)√
3(1− 3q)
≥ 2q(1− 2q)2 − 29
(1− 3q) =8
729(9q − 2)(81q2 − 63q + 13) +
46729
> 0.
The inequality is proved. Equality holds if and only if a = b = c.
♥♥♥
3