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# Inequality 30(APMO 2005)

Problem:

Let $\displaystyle x,y,z$ be positive real numbers such that $\displaystyle xyz=8$. Prove that

$\displaystyle \frac{x^2}{\sqrt{(x^3+1)(y^3+1)}}+\frac{y^2}{\sqrt{(y^3+1)(z^3+1)}}+\frac{z^2}{\sqrt{(z^3+1)(x^3+1)}}\geq\frac{4}{3}$.

Solution:

From the AM-GM ineuqality we know that

$\displaystyle \frac{1}{\sqrt{x^3+1}}=\frac{1}{\sqrt{(x+1)(x^2-x+1)}}\geq \frac{2}{(x+1)+(x^2-x+1)}=\frac{2}{x^2+2}$.

Doing that cyclic over the $\displaystyle 3$ fractions we get that

$\displaystyle \sum_{cyc}\frac{x^2}{\sqrt{(x^3+1)(y^3+1)}}\geq \frac{4x^2}{(x^2+2)(y^2+2)}$.

So, it suffices to prove that

$\displaystyle \frac{4x^2}{(x^2+2)(y^2+2)}\geq \frac{4}{3}$, or $\displaystyle 3\sum_{cyc}x^2(z^2+2)\geq \prod_{cyc}(x^2+2)$.

After expanding, the inequality is equivalent to

$\displaystyle 2\sum_{cyc}x^2+\sum_{cyc}x^2y^2\geq 72$.

But the last relation holds due to the AM-GM inequality and from the hypothesis, since $\displaystyle \sum_{cyc}x^2y^2\geq 3\sqrt[3]{8^4}=48\wedge 2\sum_{cyc}x^2\geq 6\sqrt[3]{8^2}=24$.

Adding up these $\displaystyle 2$ relations we get the desired result, Q.E.D.