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Let $x,y,z$ be real positive numbers such that $x^2+y^2+z^2=3$. Prove that :

$$\frac{x^3}{y^2+z^2}+\frac{y^3}{z^2+x^2}+\frac{z^3}{x^2+y^2} \geq \dfrac{3}{2}. $$

I try to write this expression as:

$$\frac{x^4}{x(y^2+z^2)}+\frac{y^4}{y(z^2+x^2)}+\frac{z^4}{z(x^2+y^2)}$$ and then I try to apply Cauchy-Buniakowsky but still nothing.

I need a proof/idea without derivatives.

thanks for your help.

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By Cauchy-Schwarz you want to show that $\sum x(y^2+z^2) \leq 6$. But this is equivalent to $$\begin{eqnarray}\sum x(3-x^2) \leq 6 \\ \Leftrightarrow 3\sum x \leq 6 + \sum x^3\end{eqnarray}$$ This then follows from $x^3 + 1 + 1 \ge 3x$ (AM-GM) and add up.

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The function $f(w)=w^{3/2}/(3-w)$ is convex on $(0,3)$ so $${1\over 3}\left[f(x^2)+f(y^2)+f(z^2)\right]\geq f\left([(x^2+y^2+z^2)/3]\right).$$

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  • $\begingroup$ This is nice and clean, but how do you know $f$ is convex without using derivatives? $\endgroup$ – Mike Spivey Jan 30 '13 at 21:51
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    $\begingroup$ Derivatives were forbidden after I answered. $\endgroup$ – user940 Jan 30 '13 at 21:53
  • $\begingroup$ Fair enough! (extra characters) $\endgroup$ – Mike Spivey Jan 30 '13 at 21:54
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We need to prove that $$\sum_{cyc}\frac{x^3}{3-x^2}\geq\frac{3}{2}$$ or $$\sum_{cyc}\left(\frac{x^3}{3-x^2}-\frac{1}{2}\right)\geq0$$ or $$\sum_{cyc}\frac{(x-1)(2x^2+3x+3)}{3-x^2}\geq0$$ or $$\sum_{cyc}\left(\frac{(x-1)(2x^2+3x+3)}{3-x^2}-2(x^2-1)\right)\geq0$$ or $$\sum_{cyc}\frac{(x-1)^2(2x^2+6x+3)}{3-x^2}\geq0.$$ Done!

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