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?

How about this:
$$
(\sqrt{x+y}+\sqrt{y+z}+\sqrt{z+x})^2=2(x+y+z)+2\left[\sqrt{(x+y)(y+z)}+\sqrt{(y+z)(z+x)}+\sqrt{(z+x)(x+y)}\right]\leq
2(x+y+z)+(x+2y+z)+(x+y+2z)+(2x+y+z)=6(x+y+z)
$$
where one uses the GM-AM inequality.
Finally, use $0\leq x\leq y\implies \sqrt{x}\leq\sqrt{y}$.
