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. :)