<s>[INST] 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?
 [/INST] 
We may assume that $x+y+z=1$, so we have to prove $$\sqrt{1-x}+\sqrt{1-y}+\sqrt{1-z}\leq \sqrt{6}$$

By the RMS-AM (root-mean square - arithmetic mean) inequality:
$$
\frac{\sqrt{1-x}+\sqrt{1-y}+\sqrt{1-z}}{3} \le \sqrt{\frac{\left(\sqrt{1-x}\right)^2+\left(\sqrt{1-y}\right)^2+\left(\sqrt{1-z}\right)^2}{3}} = \sqrt{\frac{2}{3}}
$$
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