<s>[INST] Proving $1/(a+b) + 1/(b+c) + 1/(c+a) > 3/(a+b+c)$ for positive $a, b, c\,$? I have to prove that:

$$ \frac{1}{a+b} + \frac{1}{b+c} + \frac{1}{c+a} > \frac{3}{a+b+c},$$
where $a, b , c$ are positive real numbers.

I am thinking about using arithmetical and geometrical averages:
$$A_{3} = \frac{a_{1}+a_{2}+a_{3}}{3},$$
$$G_{3} = \sqrt[3]{a_{1}×a_{2}×a_{3}},$$
$$A_{3}\ge G_{3}.$$
However, I am not sure how to do this.
I have tried to substract one side from the other:
$$ \frac{1}{a+b} + \frac{1}{b+c} + \frac{1}{c+a} - \frac{3}{a+b+c} > 0.$$
I have also tried to reverse both sides of inequality.
I would appreciate if you could just give me a hint.
 [/INST] By AM-HM inequality
$$\frac13 \left(\frac1{a+b}+\frac1{b+c}+\frac1{c+a}\right)\ge
\frac{3}{(a+b)+(b+c)+(c+a)}.$$
Hence $\displaystyle \frac1{a+b}+\frac1{b+c}+\frac1{c+a}\ge \frac{9}2\frac1{a+b+c}$. Because $\dfrac92>3$ the problem is solved.
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