<s>[INST] Inequality.$\sqrt{\frac{2a}{b+c}}+\sqrt{\frac{2b}{c+a}}+\sqrt{\frac{2c}{a+b}} \leq 3$ 
Let $a,b,c \gt 0$. Prove that (Using Cauchy-Schwarz) :
  $$\sqrt{\frac{2a}{b+c}}+\sqrt{\frac{2b}{c+a}}+\sqrt{\frac{2c}{a+b}} \leq 3$$

I tried to use Cauchy-Schwarz in the following form
$$\sqrt{Ax}+\sqrt{By}+\sqrt{Cz}\leq \sqrt{(A+B+C)(x+y+z)}\tag{1}.$$ 
I wrote $$\sqrt{\frac{2a}{b+c}}+\sqrt{\frac{2b}{c+a}}+\sqrt{\frac{2c}{a+b}}=\frac{\sqrt{2a(c+a)(a+b)}+\sqrt{2b(b+c)(a+b)}+\sqrt{2c(b+c)(c+a)}}{\sqrt{(a+b)(b+c)(c+a)}}.$$
and then I applied on $(1)$: 
\begin{eqnarray}
A &=& 2a(c+a)   &\mbox{and}&  x=a+b;\\ 
B &=& 2b(a+b)   &\mbox{and}&  y=b+c;\\
C &=& 2c(b+c)   &\mbox{and}&  z=c+a ,
\end{eqnarray}
but I did not obtain anything. Thanks for your help. :)
 [/INST] We will begin with an observation. Consider the function $$f(a,b,c)=\sqrt{\frac{2a}{b+c}}+\sqrt{\frac{2b}{c+a}}+\sqrt{\frac{2c}{a+b}} .$$ It has the property that $$f(ta,tb,tc)=f(a,b,c).$$ We may thus assume without loss of generality that $$a+b+c=1$$, by setting $$t=\frac{1}{a+b+c}.$$ Under this assumption, we may rewrite the function as $$\sqrt{\frac{2a}{1-a}}+\sqrt{\frac{2b}{1-b}}+\sqrt{\frac{2c}{1-c}}$$ If $a=1-\epsilon$ for $\epsilon>0$, $b+c=\epsilon$. Now note that in order for these square roots to be defined over the reals, then $0\leq a,b,c<1$, which means that $b,c<\epsilon$. Now for such an $\epsilon$,  $$\sqrt{\frac{2b}{1-b}}, \sqrt{\frac{2c}{1-c}}$$ are both close to zero, and $$\sqrt{\frac{2a}{1-a}}$$ is large.
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