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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.

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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.

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$$\sum_{cyc}\sqrt{\frac{a+b}{c}}-2\sum_{cyc}\sqrt{\frac{c}{a+b}}=\sum_{cyc}\frac{a+b-2c}{\sqrt{c(a+b)}}=$$ $$=\sum_{cyc}\frac{b-c-(c-a)}{\sqrt{c(a+b)}}=\sum_{cyc}(a-b)\left(\frac{1}{\sqrt{b(a+c)}}-\frac{1}{\sqrt{a(b+c)}}\right)=$$ $$=\sum_{cyc}\frac{(a-b)^2c}{\sqrt{ab(a+c)(b+c)}\left(\sqrt{a(b+c)}+\sqrt{b(a+c)}\right)}\geq0.$$ Done!

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