Prove this inequality


by utkarshakash
Tags: inequality, prove
utkarshakash
utkarshakash is offline
#1
Nov20-12, 08:12 AM
P: 638
1. The problem statement, all variables and given/known data
If a,b,c are the positive real numbers, prove that [itex]a^2(1+b^2)+b^2(1+c^2)+c^2(1+a^2) \geq 6abc[/itex]

2. Relevant equations

3. The attempt at a solution
With a little simplification L.H.S = [itex](a^2+b^2+c^2)+(a^2b^2+b^2c^2+c^2a^2)[/itex]
Using A.M>=G.M
[itex]\dfrac{a^2+b^2+c^2}{3} \geq (a^2b^2c^2)^{\frac{1}{3}} \\
a^2+b^2+c^2 \geq 3a^{2/3}b^{2/3}c^{2/3} \\
[/itex]
Also
[itex] \dfrac{a^2b^2+b^2c^2+c^2a^2}{3} \geq (a^2b^2.b^2c^2.c^2a^2)^{1/3} \\
a^2b^2+b^2c^2+c^2a^2 \geq 3a^{4/3}b^{4/3}c^{4/3}
[/itex]
Adding the two inequalities
[itex]
(a^2+b^2+c^2)+(a^2b^2+b^2c^2+c^2a^2) \geq 3[a^{2/3}b^{2/3}c^{2/3}+a^{4/3}b^{4/3}c^{4/3}]
[/itex]

Now how do I simplify next?
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mfb
mfb is offline
#2
Nov20-12, 08:24 AM
Mentor
P: 10,824
Your last inequality can be written as
LHS >= 3 (x+x^2)
with an appropriate x.

And you have to show that
LHS >= 6x3/2

You can just use the same trick again at your new sum.
utkarshakash
utkarshakash is offline
#3
Nov20-12, 08:38 AM
P: 638
Quote Quote by mfb View Post
Your last inequality can be written as
LHS >= 3 (x+x^2)
with an appropriate x.

And you have to show that
LHS >= 6x3/2

You can just use the same trick again at your new sum.
Thanks. I got it.


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