Let $a,b,c$ are the lengths of a triangle. Prove that: $\frac{\sqrt{a+b-c}}{\sqrt{a}+\sqrt{b}-\sqrt{c}}+\frac{\sqrt{b+c-a}}{\sqrt{b}+\sqrt{c}-\sqrt{a}}+\frac{\sqrt{c+a-b}}{\sqrt{c}+\sqrt{a}-\sqrt{b}}\leq 3$
I tried Schur inequality:
It can be written as:
$\sum_{cyc}^{}\sqrt{a+b-c}(\sqrt{b}+\sqrt{c}-\sqrt{a})(\sqrt{c}+\sqrt{a}-\sqrt{b})\leq 3\prod _{cyc}(\sqrt{a}+\sqrt{b}-\sqrt{c})$ which is very complicated and opposite from the one I knew is:$xyz\geq (x+y-z)(y+z-x)(z+x-y),\forall x,y,z\geq 0$
Also I do the another idea: Suppose: $a+b+c=3$ so the inequality can be written as: $\sum_{cyc}^{}\frac{\sqrt{3-2c}}{\sqrt{a}+\sqrt{b}-\sqrt{c}}\leq 3$
but this step doesn't make the denominators easier and it still stuck here :()
Can anyone make the easier solution with Schur inequality? Thank you!