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$a,b,c \in \mathbb{R^+} \text{such that }a+b+c=2$. Prove inequality $$\frac a{ab+2c}+\frac b{bc+2a}+\frac c{ca+2b} \ge \frac 98$$

I tried

  • $$LHS = \sum \frac{1}{b+2\cdot c/a} \ge \frac 9 {2+2(\sum c/a)} \longrightarrow failed$$
  • $$\frac a {ab+2c} \ge \frac 9{16}a \longrightarrow failed$$
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1 Answer 1

up vote 15 down vote accepted

\begin{align} \sum_{cyc}{\frac{a}{ab+2c}}=\sum_{cyc}{\frac{a}{ab+(a+b+c)c}}=\sum_{cyc}{\frac{a}{(a+c)(b+c)}}& =\frac{\sum_{cyc}{a(a+b)}}{(a+b)(a+c)(b+c)} \\ & =\frac{a^2+b^2+c^2+ab+ac+bc}{(a+b)(a+c)(b+c)} \\ & \geq \frac{\frac{2}{3}(a+b+c)^2}{(a+b)(a+c)(b+c)} \\ & \geq \frac{\frac{2}{3}(a+b+c)^2}{\left(\frac{(a+b)+(a+c)+(b+c)}{3}\right)^3} \\ & =\frac{9}{8} \end{align}

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Ivan, you have my respect :-) – Iuli Feb 15 '13 at 12:50
Congrats, and +1. – 1015 Feb 15 '13 at 13:25

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