Prove that: $\sqrt{x+y} + \sqrt{y+z} + \sqrt{z+x} \leq \sqrt{6(x+y+z)}$ Prove that for nonegative $x,y,z$ we have: $$\sqrt{x+y} + \sqrt{y+z} + \sqrt{z+x} \leq \sqrt{6(x+y+z)}$$

I prove that using the tangent line method. We may assume that $x+y+z=1$, so you we have to prove $$\sqrt{1-x}+\sqrt{1-y}+\sqrt{1-z}\leq \sqrt{6}$$
A tangent on $f(x)=\sqrt{1-x}$ at $x={1\over 3}$ is $$y=-{\sqrt{6}\over 4}x+{5\sqrt{6}\over 12}$$
So we have, for all $x\in[0,1]$:
$$\sqrt{1-x} \leq -{\sqrt{6}\over 4}x+{5\sqrt{6}\over 12}$$  and we are done...
I wonder if there is elegant method avoiding calculus?

Hint: This is just
$$
(a+b+c)^2\leq 3(a^2+b^2+c^2)\iff ab+bc+ca\leq a^2+b^2+c^2
$$
which is proved easily enough. You can start by $a\equiv\sqrt{x+y}$, $b\equiv\ldots$, etc.
