Prove that $\sum\limits_{cyc}\sqrt{\frac{a+b}{c}}\ge2\sum\limits_{cyc}\sqrt{\frac{c}{a+b}}$ Let $a,b,c$ be positive numbers. Then we need to prove
$\sqrt{\frac{a+b}{c}}+\sqrt{\frac{b+c}{a}}+\sqrt{\frac{c+a}{b}}\ge2\left(\sqrt{\frac{c}{a+b}}+\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{c+a}}\right).$
I have an idea to set $x=\frac a{b+c}$, $y=\frac b{c+a},z=\frac c{a+b}$ then
$\frac1{1+x}+\frac1{1+y}+\frac1{1+z}=2$ 
and we need to prove 
$\frac1{\sqrt x}+\frac1{\sqrt y}+\frac1{\sqrt z}\ge2\left(\sqrt x+\sqrt y+\sqrt z\right)$  
But I could not go further.
 A: It is a consequence of Chebychev's inequality:
$$
\sum_{cyc}\sqrt{\frac{a+b}{c}}≥2\sum_{cyc}\sqrt{\frac{c}{a+b}}\iff\sum_{cyc}\frac{a+b-2c}{\sqrt{c(a+b)}}≥0
$$
Since the $a+b-2c$ and $\frac{1}{\sqrt{c(a+b)}}$ are ordered in the same way, we can apply Chebychev's inequality to obtain:
$$
\sum_{cyc}\frac{a+b-2c}{\sqrt{c(a+b)}}≥\frac{1}{3}\left(\sum_{cyc}a+b-2c\right)\left(\sum_{cyc}\frac{1}{\sqrt{c(a+b)}}\right)=0
$$
Edit:
In case you are not familiar with this approach:
If we consider two real sequences $a_1,a_2,…,a_n$ and $b_1,b_2,…,b_n$ for which $a_1≤a_2≤…≤a_n$ and $b_1≤b_2≤…≤b_n$, then Chebyvhev tells us, that:
$$
\frac{a_1b_1+a_2b_2+…+a_nb_n}{n}≥\frac{a_1+a_2+…+a_n}{n}\cdot\frac{b_1+b_2+…+b_n}{n}
$$
We are allowed to use it in this case, because by symmetry, we can assume $a≥b≥c>0$. This implies:
$$
a+b-2c≥a+c-2b≥b+c-2a
$$
And:
$$
ab≥ac≥bc\iff a(b+c)≥b(a+c)≥c(a+b) \iff\\ \frac{1}{\sqrt{c(a+b)}}≥\frac{1}{\sqrt{b(a+c)}}≥\frac{1}{\sqrt{a(b+c)}}
$$
So the two sequences we have in the above inequality are indeed ordered in the same way.
A: $$\sum_{cyc}\sqrt{\frac{a+b}{c}}-2\sum_{cyc}\sqrt{\frac{c}{a+b}}=\sum_{cyc}\frac{a+b-2c}{\sqrt{c(a+b)}}=$$
$$=\sum_{cyc}\frac{b-c-(c-a)}{\sqrt{c(a+b)}}=\sum_{cyc}(a-b)\left(\frac{1}{\sqrt{b(a+c)}}-\frac{1}{\sqrt{a(b+c)}}\right)=$$
$$=\sum_{cyc}\frac{(a-b)^2c}{\sqrt{ab(a+c)(b+c)}\left(\sqrt{a(b+c)}+\sqrt{b(a+c)}\right)}\geq0.$$
Done!
