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} $

Question

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.

Answer 1

Since $\sqrt{x}$ is concave down, Jensen's inequality tells us that
$ \dfrac 12 ( \sqrt{2x} + \sqrt{2y}) \leq \sqrt{ \dfrac{ 2x + 2y } 2 } = \sqrt{x+y}$.
Summing cyclically gives the desired result.