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$a;b;c\in \mathbb{R}^{+}$ such that $abc=1$ Prove : $\frac{a}{ac+1}+ \frac{b}{ab+1}+ \frac{c}{bc+1} \leq \frac{1}{2}(a^2+b^2+c^2)$

Lê Tấn Khang
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    Have you tried multiplying? They're all positive variables so you can multiply both sides of the inequality by $2(ac+1)(ab+1)(bc+1)$ and expand both sides to see what you get. Once you've done this put your solution - or attempted solution up. –  Oct 24 '13 at 06:23
  • If you have a new question, please post it rather than editing an old post. –  Oct 24 '13 at 07:13
  • I'm sorry @Sanchez !!! – Lê Tấn Khang Oct 24 '13 at 08:16

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$\frac{a}{ac+1} \leq \frac{a}{2\sqrt{ac}} = \frac{1}{2} a \sqrt{b}$. So it suffices to show that $$a^2+b^2+c^2 \ge (abc)^{1/6} (a\sqrt{b} + b\sqrt{c} + c\sqrt{a}) = \sum_{cyc} a^{7/6}b^{2/3}c^{1/6}$$ This follows from adding up the AM-GM inequalities: $$7a^2+4b^2+c^2 \ge 12 a^{7/6}b^{2/3}c^{1/6}$$ and its cyclic variants.