# Inequality. $\sqrt{\frac{ab+bc+ca}{3}} \leq \sqrt[3]{\frac{(a+b)(b+c)(c+a)}{8}}$

Let $a,b,c$ be nonnegative reals. Prove that :

$$\sqrt{\frac{ab+bc+ca}{3}} \leq \sqrt[3]{\frac{(a+b)(b+c)(c+a)}{8}}.$$

Thanks!

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It is clearly equivalent to prove that $$(a+b)^2(b+c)^2(c+a)^2 \ge \frac{64}{27} (ab + bc + ca)^3$$ We prove, more strongly, that $$(a+b)^2(a+c)^2(b+c)^2 \ge \frac{64}{81} (a+b+c)^2(ab+bc+ca)^2$$ The inequality then follows by noticing that $(a+b+c)^2 \ge 3(ab+bc+ca)$
To prove the inequality, note first, that $(a+b)(b+c)(c+a) \ge 2 \sqrt{ab} \cdot 2 \sqrt{bc} \cdot 2 \sqrt{ca} = 8abc$. Now, \begin{align}(a+b)(b+c)(c+a) &= (a+b+c)(ab+bc+ca) - abc \\ &\ge (a+b+c)(ab+bc+ca) - \frac 18 (a+b)(b+c)(c+a) \end{align} $$\implies (a+b)(b+c)(c+a) \ge \frac{8}{9} (a+b+c)(ab+bc+ca)$$ Squaring both sides gives us the desired.
you use two notation : $,b,c$ and $x,y,z$. – Iuli Sep 12 '12 at 9:29