How can I approach this inequality? Let $a, b$ and $c$ be three non-zero positive numbers. Show that:
$$\sqrt{\frac{2a}{a + b}} + \sqrt{\frac{2b}{b + c}} + \sqrt{\frac{2c}{a + c}} \leq 3$$
I know the triangular inequality would help here, but I don't know how to approach it.
I started by $a+b≥a$ then that gives $\frac{1}{a+b}≤\frac{1}{a}$ by muliplyting both sides by $2a$ we get $\frac{2a}{a+b}≤\frac{2a}{a}$ which leads eventually to $\frac{2a}{a+b}≤2$ and by adding the square root to both sides we get $\sqrt{\frac{2a}{a+b}}\leq\sqrt2$ and doing the same thing to the other terms we get $\sqrt{\frac{2a}{a+b}}+\sqrt{\frac{2b}{b+c}}+\sqrt{\frac{2c}{c+a}}\leq3\sqrt2$  beyond that I don't have any idea if that would lead to anything useful or not.

By C-S $$\sum_{cyc}\sqrt{\frac{2a}{a+b}}\leq\sqrt{2\sum_{cyc}\frac{a}{(a+b)(a+c)}\sum_{cyc}(a+c)}=$$
$$=\sqrt{\frac{8(ab+ac+bc)(a+b+c)}{\prod\limits_{cyc}(a+b)}}\leq3,$$ where the last inequality it's just $$\sum_{cyc}c(a-b)^2\geq0.$$
