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.

This is not the most elegant approach, but since the inequality is homogeneous we may as well assume $a=x>0, b=1, c=y>0$ and study the behaviour of 
$$ f(x,y) = \sqrt{\frac{2x}{x+1}}+\sqrt{\frac{2}{1+y}}+\sqrt{\frac{2y}{x+y}} $$
over $(0,+\infty)^2$. If $x\to 0$ or $y\to 0$ we have $f(x,y)\leq 2\sqrt{2}<3$. By solving
$\frac{\partial f}{\partial x}=0$ we find that the first partial derivative only vanishes over the curve $y=x^2$ and over the curve $y=\frac{1}{2}\left(-3x-x^2+(1+x)\sqrt{4x+x^2}\right)$. By solving $\frac{\partial f}{\partial y}=0$ we find three curves, one of them being $y=\sqrt{x}$ and the other two being defined only for $x\geq 4$. The only point in which an $f_x$-curve meets an $f_y$-curve is $(1;1)$, so $f(1,1)=3$ is the only stationary point and an actual maximum.
