<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] Since $a > 0$, $b > 0$, and $c > 0$, we have
$$ 0 < a +b < a+b + c, $$
and therefore upon dividing both sides of this inequality by $(a+b)(a+b+c) > 0$, we get
$$
0 < \frac{1}{a+b+c} < \frac{1}{a+b},
$$
which implies
$$
\frac{1}{a+b} > \frac{1}{a+b+c}.  \tag{1}
$$
Similarly, we have the inequalities
$$
\frac{1}{b+c} > \frac{1}{a+b+c}  \tag{2}
$$
and
$$
\frac{1}{c+a} > \frac{1}{a+b+c}.  \tag{3}
$$
Using (1), (2), and (3), we obtain
\begin{align}
\frac{1}{a+b} + \frac{1}{b+c} + \frac{1}{c+a } &> \frac{1}{a + b + c} + \frac{1}{a + b + c} + \frac{1}{a + b + c} \\
&= \frac{3}{a + b + c},
\end{align}
as required.
</s>