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For positive numbers $a$,$b$,$c$ , with $abc=1$ prove $$\frac{1}{a+b} +\frac{1}{a+c}+\frac{1}{c+b} \leq\frac{3}{2}$$

Souvik Dey
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yibotg
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    Is this a homework question? Can you show us what have you tried so far? – Marra Apr 24 '13 at 03:12
  • No,To prove another inequality For $\lambda\geq 1$ ,$\sum\frac{1}{a+b+\lambda} \leq \frac{3}{2+\lambda}$ . I got the above inequality. – yibotg Apr 24 '13 at 03:38

1 Answers1

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The problem as stated is false. Take $a=b=0.1$, $c=100$. $\frac{1}{a+b}=5$.

However, if $\min(a,b,c)\ge 1$ (for example, if they are integers), then it's true, by the other posted solution.

vadim123
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