Nesbitt by Nesbitt The title not says I'm Nesbitt but just only says it's a refinement of Nesbitt's inequality by itself so we have :

Let $a,b,c>0$ and $a\geq b \geq c$ then we have :
  $$\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\geq\frac{\sqrt{a}}{\sqrt{b}+\sqrt{c}}+\frac{\sqrt{b}}{\sqrt{a}+\sqrt{c}}+\frac{\sqrt{c}}{\sqrt{b}+\sqrt{a}}$$

I think (I have tested numerically ) we can use majorization like this (with $a,b,c>0$ and $a\geq b \geq c$):

First line of the majorization 

$$\frac{a}{b+c}+\frac{b}{a+c}\geq \frac{\sqrt{a}}{\sqrt{b}+\sqrt{c}}+\frac{\sqrt{b}}{\sqrt{a}+\sqrt{c}}$$

Second line of the majorization 

$$\Big(\frac{a}{b+c}+\frac{b}{a+c}\Big)\Big(\frac{a}{b+c}+\frac{c}{a+b}\Big)\geq \Big(\frac{\sqrt{a}}{\sqrt{b}+\sqrt{c}}+\frac{\sqrt{b}}{\sqrt{a}+\sqrt{c}}\Big)\Big(\frac{\sqrt{a}}{\sqrt{b}+\sqrt{c}}+\frac{\sqrt{c}}{\sqrt{b}+\sqrt{a}}\Big)$$

Third line of the majorization 

$$\prod_{cyc}\Big(\frac{a}{b+c}+\frac{b}{a+c}\Big)\geq\prod_{cyc}\Big(\frac{\sqrt{a}}{\sqrt{b}+\sqrt{c}}+\frac{\sqrt{b}}{\sqrt{a}+\sqrt{c}}\Big)$$
But this way's is too long so I forget this one .
I have tried also the Transformation of Ravi but without success .
So if you have a nice idea I take !
Thanks a lot for sharing your time and knowledge .

We need to prove that: $$\sum_{cyc}\frac{a^2}{b^2+c^2}\geq\sum_{cyc}\frac{a}{b+c},$$where a, b and c are positives. 
Indeed, $$\sum_{cyc}\frac{a^2}{b^2+c^2}-\sum_{cyc}\frac{a}{b+c}=\sum_{cyc}\frac{ab(a-b)-ac(c-a)}{(b^2+c^2)(b+c)}=$$
$$=\sum_{cyc}(a-b)\left(\frac{ab}{(b^2+c^2)(b+c)}-\frac{ab}{(c^2+a^2)(c+a)}\right)=$$
$$=\sum_{cyc}\tfrac{(a-b)^2ab(a^2+b^2+c^2+ab+ac+bc)}{(b^2+c^2)(b+c)(c^2+a^2)(c+a)}\geq0.$$
