<s>[INST] Inequality using Cauchy-Schwarz Let $a,b,c\in\mathbb{R}^+$, prove that $$\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{c+a}}+\sqrt{\frac{c}{a+b}}\leq \sqrt{\frac{3}{2}\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)} $$
Hi everyone, I've been trying to do this exercise but any method that I tried has failed. First I tried to use $v=(\sqrt{\frac{a}{b+c}},\sqrt{\frac{b}{c+a}},\sqrt{\frac{c}{a+b}})$ and $w=(1,1,1)$ (the most obvious) but didn't work. Then I did this $$LHS= \frac{\sqrt{a(c+a)(a+b)}+\sqrt{b(b+c)(a+b)}+\sqrt{c(b+c)(c+a)}}{\sqrt{(b+c)(c+a)(a+b)}}\leq \frac{\sqrt{(a(a+b)+b(b+c)+c(c+a))(c+a+a+b+b+c)}}{\sqrt{(b+c)(c+a)(a+b)}} = \frac{\sqrt{2((a+b+c)^2-(ab+bc+ac))(a+b+c)}}{\sqrt{(b+c)(c+a)(a+b)}}$$
Using the fact that $\sqrt{ax}+\sqrt{by}+\sqrt{cz} \leq \sqrt{(a+b+c)(x+y+z)}$ but I don't know if I'm not looking something. If someone can give me a hint I would really appreciate it
 [/INST] Just a hint. You can denote:   
$x=\frac{a}{b}, y=\frac{b}{c}, z=\frac{c}{a}$
Then we have: $xyz=1$ i.e. $z = \frac{1}{xy}$      
Then use that:  $\sqrt \frac{a}{b+c} = \sqrt \frac{1}{b/a+c/a} = \sqrt \frac{1}{1/x + z} $ and do analogically for the other terms in the LHS.
Then if you use that $z = \frac{1}{xy}$ it boils down to an inequality containing just $x$ and $y$.  
Let me see if something good will come out of this, I think it will.   
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