I have this exercise from a book I have been making my way through.
Let x,y,z be positive real numbers. Prove the following: $$\sqrt{\frac{x+y}{x+y+z}} +\sqrt{\frac{x+z}{x+y+z}} + \sqrt{\frac{y+z}{x+y+z}} \le \sqrt{6}$$
I thought that I may be able to use Cauchy's inequality to answer this question, but after a few attempts I'm not getting anywhere meaningful. I know that since clearly $x + y < x + y + z$, then $\sqrt{\frac{x+y}{x+y+z}}$ must be less than one. I tried repeating with the other fractions and proved that the once the fractions are summed up it must be less than three, but that's all I've managed to do.
Any help would be appreciated. Thanks
edit: Turns out you can use Cauchy's inequality to solve it. Let $b_1, b_2, b_3 = 1$. Then let $a_1 = \sqrt{\frac{x+y}{3(x+y+z)}}$ $a_2 = \sqrt{\frac{x+z}{3(x+y+z)}}$ $a_3 = \sqrt{\frac{y+z}{3(x+y+z)}}$ This leads to the solution.
Best Answer
It's a convexity result. Let : $$a = \frac{x+y}{(x+y+z)}$$ $$b = \frac{x+z}{(x+y+z)}$$ $$c= \frac{y+z}{(x+y+z)}$$ The square root is concave : $$\sqrt{\frac23}=\sqrt{\frac13 a + \frac13 b + \frac13 c} \geq \frac13 \Big(\sqrt{a} + \sqrt{b} + \sqrt{c}\Big)$$ Which gives the result.
EDIT : Concavity of $f$ implies that $f$ is above every function : $$\Delta_y(x) = \frac{f(x) - f(y)}{x - y}$$