If $a,b,c$ be the sides of a triangle, then prove that$$\frac{\sqrt{b+c-a}}{\sqrt{b}+\sqrt{c}-\sqrt{a}}+\frac{\sqrt{c+a-b}}{\sqrt{c}+\sqrt{a}-\sqrt{b}}+\frac{\sqrt{a+b-c}}{\sqrt{a}+\sqrt{b}-\sqrt{c}}\leq 3$$
My Attempt:
I tried by putting $a=x+y,b=y+z,c=z+x$ where $x,y,z>0$ but got stuck
Let $a\geq b\geq c$.
Thus, $$3-\sum_{cyc}\frac{\sqrt{b+c-a}}{\sqrt{b}+\sqrt{c}-\sqrt{a}}=\sum_{cyc}\frac{\sqrt{b}+\sqrt{c}-\sqrt{a}-\sqrt{b+c-a}}{\sqrt{b}+\sqrt{c}-\sqrt{a}}=$$ $$=\sum_{cyc}\frac{\left(\sqrt{b}+\sqrt{c}\right)^2-\left(\sqrt{a}+\sqrt{b+c-a}\right)^2}{\left(\sqrt{b}+\sqrt{c}-\sqrt{a}\right)\left(\sqrt{b}+\sqrt{c}+\sqrt{a}+\sqrt{b+c-a}\right)}=$$ $$=2\sum_{cyc}\frac{\sqrt{bc}-\sqrt{a(b+c-a)}}{\left(\sqrt{b}+\sqrt{c}-\sqrt{a}\right)\left(\sqrt{b}+\sqrt{c}+\sqrt{a}+\sqrt{b+c-a}\right)}=$$ $$=2\sum_{cyc}\frac{bc-a(b+c-a)}{\left(\sqrt{b}+\sqrt{c}-\sqrt{a}\right)\left(\sqrt{b}+\sqrt{c}+\sqrt{a}+\sqrt{b+c-a}\right)\left(\sqrt{bc}+\sqrt{a(b+c-a)}\right)}=$$ $$=2\sum_{cyc}\frac{(a-b)(a-c)}{\left(\sqrt{b}+\sqrt{c}-\sqrt{a}\right)\left(\sqrt{b}+\sqrt{c}+\sqrt{a}+\sqrt{b+c-a}\right)\left(\sqrt{bc}+\sqrt{a(b+c-a)}\right)}\geq$$ $$\geq\tfrac{2(a-b)(a-c)}{\left(\sqrt{b}+\sqrt{c}-\sqrt{a}\right)\left(\sqrt{b}+\sqrt{c}+\sqrt{a}+\sqrt{b+c-a}\right)\left(\sqrt{bc}+\sqrt{a(b+c-a)}\right)}+\tfrac{2(b-a)(b-c)}{\left(\sqrt{a}+\sqrt{c}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{c}+\sqrt{b}+\sqrt{a+c-b}\right)\left(\sqrt{ac}+\sqrt{b(a+c-b)}\right)}=$$ $$=2(a-b)\left(\tfrac{a-c}{\left(\sqrt{b}+\sqrt{c}-\sqrt{a}\right)\left(\sqrt{b}+\sqrt{c}+\sqrt{a}+\sqrt{b+c-a}\right)\left(\sqrt{bc}+\sqrt{a(b+c-a)}\right)}-\tfrac{b-c}{\left(\sqrt{a}+\sqrt{c}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{c}+\sqrt{b}+\sqrt{a+c-b}\right)\left(\sqrt{ac}+\sqrt{b(a+c-b)}\right)}\right)\geq0$$ because $$a-c\geq b-c,$$ $$\sqrt{a}+\sqrt{c}-\sqrt{b}\geq \sqrt{b}+\sqrt{c}-\sqrt{a},$$ $$\sqrt{a}+\sqrt{c}+\sqrt{b}+\sqrt{a+c-b}\geq \sqrt{a}+\sqrt{c}+\sqrt{b}+\sqrt{b+c-a}$$ and $$\sqrt{ac}+\sqrt{b(a+c-b)}\geq \sqrt{bc}+\sqrt{a(b+c-a)}.$$ Done!