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.

Other solution is Cauchy-Schwarz, that can be applied in one of its several versions.

*

*This is the standard one:

$[(\frac{1}{\sqrt{a+b}})^2+(\frac{1}{\sqrt{b+c}})^2+(\frac{1}{\sqrt{c+a}})^2][((\sqrt{a+b})^2+(\sqrt{b+c})^2+(\sqrt{c+a})^2]\ge 3^2$
From here the factor $9/2$, as already found, follows.

*

*Titu's inequality https://brilliant.org/wiki/titus-lemma/#
Here we directly have:
$\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\ge\frac{(1+1+1)^2}{2(a+b+c)}$

*

*Holders form: https://brilliant.org/wiki/holders-inequality/ :

$(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a})^{1/2}((a+b)+(b+c)+(c+a))^{1/2}\ge 1^{1/2}+1^{1/2}+1^{1/2}$
I wrote this answer also for a reference of mine to the different forms of C.S.
