If a,b,c are sides of a triangle, prove: $ \sqrt{a+b-c} + \sqrt{b+c-a} + \sqrt{c+a-b} \le \sqrt{a} + \sqrt{b} + \sqrt{c} $ I did substitute $a=x+y, b=x+z, c=y+z$ and I arrived at $\sqrt{2x} + \sqrt{2y} + \sqrt{2z} \le \sqrt{x+y} + \sqrt{x+z} + \sqrt{y+z}$.
However, after this, I tried various methods like AM-GM and Cauchy-Schwarz inequality for hours and I still can't prove it. Can someone help please? Thanks.

Let $a\geq b\geq c$ and $f(x)=\sqrt{x}$.
Hence, $a+b-c\geq a+c-b\geq b+c-a$, $(a+b-c,a+c-b,b+c-a)\succ(a,b,c)$ and since $f$ is a concave function, the starting inequality it's just Karamata.
Done!
