# Prove this inequality

1. Nov 20, 2012

### utkarshakash

1. The problem statement, all variables and given/known data
If a,b,c are the positive real numbers, prove that $a^2(1+b^2)+b^2(1+c^2)+c^2(1+a^2) \geq 6abc$

2. Relevant equations

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

Now how do I simplify next?

2. Nov 20, 2012

### Staff: Mentor

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.

3. Nov 20, 2012

### utkarshakash

Thanks. I got it.