How to Prove an Inequality Involving Positive Real Numbers?

[utkarshakash]

Homework Statement

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

Homework Equations

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}} \\<br /> a^2+b^2+c^2 \geq 3a^{2/3}b^{2/3}c^{2/3} \\<br />
Also
\dfrac{a^2b^2+b^2c^2+c^2a^2}{3} \geq (a^2b^2.b^2c^2.c^2a^2)^{1/3} \\<br /> a^2b^2+b^2c^2+c^2a^2 \geq 3a^{4/3}b^{4/3}c^{4/3}<br />
Adding the two inequalities
<br /> (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}]<br />

Now how do I simplify next?

 

[mfb]

Your last inequality can be written as
LHS >= 3 (x+x^2)
with an appropriate x.

And you have to show that
LHS >= 6x^3/2

You can just use the same trick again at your new sum.

 

[utkarshakash]

Your last inequality can be written as
LHS >= 3 (x+x^2)
with an appropriate x.

And you have to show that
LHS >= 6x^3/2

You can just use the same trick again at your new sum.

Thanks. I got it.