Prove this inequality

by utkarshakash
Tags: inequality, prove
 P: 542 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 $\dfrac{a^2b^2+b^2c^2+c^2a^2}{3} \geq (a^2b^2.b^2c^2.c^2a^2)^{1/3} \\ a^2b^2+b^2c^2+c^2a^2 \geq 3a^{4/3}b^{4/3}c^{4/3}$ Adding the two inequalities $(a^2+b^2+c^2)+(a^2b^2+b^2c^2+c^2a^2) \geq 3[a^{2/3}b^{2/3}c^{2/3}+a^{4/3}b^{4/3}c^{4/3}]$ Now how do I simplify next?
 Mentor P: 10,514 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.
P: 542
 Quote by mfb 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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