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.
 A: 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.
A: 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 $$\sum_{cyc}c(a-b)^2\geq0.$$
A: It's not a clean solution but it works. Consider the change of variable $0<x:=\frac{b}{a}, 0<y:=\frac{c}{b}, 0 < x :=\frac{a}{c} $. Observe that $xyz=1$. Now the inequality reads 
$$f(x,y,z)=\sqrt{\frac{2}{1+x}}+\sqrt{\frac{2}{1+y}}+\sqrt{\frac{2}{1+z}}\leq 3$$
That means we have to find de maxima of $f$ subject to restriction $g(x,y,z):=xyz=1$. Now we calculate the gradients 
$$\nabla f(x,y,z)= -\sqrt{2}\left(\frac{1}{(1+x)^{3/2}},\frac{1}{(1+y)^{3/2}},\frac{1}{(1+z)^{3/2}} \right).$$
$$ \nabla g(x,y,z) = \lambda  (yz,xz,xy)$$
Then we solve $\nabla f= \lambda \nabla g$. Now we solve a system of equations.
\begin{align}
\frac{1}{(1+x)^{3/2}} & = \lambda yz \quad (1) \\
\frac{1}{(1+y)^{3/2}} & = \lambda xz \quad (2)\\
\frac{1}{(1+z)^{3/2}} & = \lambda xy \quad (3)\\
1 & = xyz \quad (4) \\
\end{align}
Observe by right side of these equations $\lambda,x,y,z \neq 0 $. Dividing (1) with (2) and $(1)$ with $(3)$, in addition with some calculations we get 
\begin{align}
h(x) &= h(y) \\
h(x) &= h(z). \\
\end{align} 
Where $h(x) := \frac{x}{(1+x)^{3/2}}$ which satisfies $h''(x)\leq 0$ in particular is concave in $[0,\infty[$ so it doesn't take the same value more than two times,that means two of the $x,y,z$ are equal. Hence, without lost of generality $x=y$. With this 
\begin{align*}
\lambda  &= \frac{x}{(1+x)^{3/2}} \\
\frac{1}{(1+z)^{3/2}} &= \lambda x^2 \\
z & = 1/x^2
\end{align*}
doing the substitutions and algebraic manipulations we get $x^2=1$. So $x=y=z=1$,
Hence the maximum is  $f(1,1,1)=3$.
