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

I finally understood what tong_nor meant so I want to write it in details.
So treat this answer as a more detailed version of the tong_nor answer.   
The Cauchy-Schwarz inequality gives us that:     
(1) $(a_1^2 + a_2^2 + a_3^2)(b_1^2 + b_2^2 + b_3^2) \geq (a_1b_1 + a_2b_2 + a_3b_3)^2$  
If we put here $a_i = \sqrt {a_i}$, $b_i = \sqrt {b_i}$ we get:    
(2) $(a_1 + a_2 + a_3)(b_1 + b_2 + b_3) \geq (\sqrt {a_1b_1} + \sqrt{a_2b_2} + \sqrt{a_3b_3})^2$    
Now in (2) we put:   
$a_1 = a$, $a_2 = b$, $a_3 = c$
$b_1 = \frac{1}{b+c}$, $b_2 = \frac{1}{c+a}$, $b_3 = \frac{1}{a+b}$    
So we get:   
(3) $(a+b+c) (\frac{1}{b+c} + \frac{1}{c+a} + \frac{1}{a+b}) \geq (\sqrt {\frac{a}{b+c}} + \sqrt {\frac{b}{c+a}} + \sqrt {\frac{c}{a+b}})^2$    
We take square root in (3) and reverse the sides:    
(4) $ \sqrt {\frac{a}{b+c}} + \sqrt {\frac{b}{c+a}} + \sqrt {\frac{c}{a+b}} \leq \sqrt {(a+b+c) (\frac{1}{b+c} + \frac{1}{c+a} + \frac{1}{a+b}) } $    
Now back to (2) where we put $a_i = \frac{x_i^2}{y_i}$ and $b_i = y_i$    
So we get:   
(5) $(\frac{x_1^2}{y_1} + \frac{x_2^2}{y_2} + \frac{x_3^2}{y_3}) (y_1+y_2+y_3)\geq (x_1+x_2 + x_3)^2$  
Now in (5) we put:   
$x_1 = a, x_2 = b, x_3 = c$
$y_1 = ab, y_2 = bc, y_3 = ca$     
and we obtain the following:    
(6) $ ( \frac{a}{b} + \frac{b}{c} + \frac{c}{a} ) (ab+bc+ca) \geq (a+b+c)^2  $   i.e.  
(7) $ ( \frac{a}{b} + \frac{b}{c} + \frac{c}{a} )  \geq \frac{(a+b+c)^2}{(ab+bc+ca)}  $    
Now we will prove that:     
(8) $(\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b}) \le \frac{3}{2}\frac{a+b+c}{ab+bc+ca}$
Let's rewrite the inequality (8). We multiply both sides by $ab+bc+ac$. Then:   
$$LHS=\frac{ab+bc+ac}{b+c}+\frac{ab+bc+ac}{c+a}+\frac{ab+bc+ac}{a+b} = \frac{a(b+c)+bc}{b+c}+\frac{b(a+c)+ac}{a+c}+\frac{c(b+a)+ab}{a+b} = a+\frac{bc}{b+c}+b+\frac{ac}{c+a}+c+\frac{ab}{a+b}$$
Now we subtract $(a+b+c)$ from both sides. This shows that:
$$(8) \Leftrightarrow \frac{bc}{b+c}+\frac{ac}{a+c}+\frac{ab}{a+b}\leq \frac{1}{2}(a+b+c) $$   
Taking into account that $4ab\leq (a+b)^2$ we get:     
$$\frac{bc}{b+c}+\frac{ac}{a+c}+\frac{ab}{a+b}\leq \frac{(b+c)^2}{4(b+c)}+\frac{(c+a)^2}{4(c+a)}+\frac{(a+b)^2}{4(a+b)} = \frac{1}{2}(a+b+c) $$
OK, this completes the proof of (8).   
So now from (4) and by using (8) and then (7) we get:
(9)  $ \sqrt {\frac{a}{b+c}} + \sqrt {\frac{b}{c+a}} + \sqrt {\frac{c}{a+b}} \leq \sqrt {(a+b+c) \cdot \frac{3}{2}\frac{(a+b+c)}{ab+bc+ca}} \leq \sqrt{\frac{3}{2}(\frac{a}{b}+\frac{b}{c}+\frac{c}{a})}$       
That's all we wanted to prove.   
