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?
 A: 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.
A: 
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:
$$
\frac{\sqrt{1-x}+\sqrt{1-y}+\sqrt{1-z}}{3} \le \sqrt{\frac{\left(\sqrt{1-x}\right)^2+\left(\sqrt{1-y}\right)^2+\left(\sqrt{1-z}\right)^2}{3}} = \sqrt{\frac{2}{3}}
$$
A: 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 $0\leq x\leq y\implies \sqrt{x}\leq\sqrt{y}$.
A: By your idea:
Let $x+y+z=3$.
Thus, we need to prove that
$$\sum_{cyc}\sqrt{3-x}\leq3\sqrt2$$ or
$$\sum_{cyc}\left(\sqrt2-\sqrt{3-x}\right)\geq0$$ or
$$\sum_{cyc}\frac{x-1}{\sqrt2+\sqrt{3-x}}\geq0$$ or
$$\sum_{cyc}\left(\frac{x-1}{\sqrt2+\sqrt{3-x}}-\frac{x-1}{2\sqrt2}\right)\geq0$$ or
$$\sum_{cyc}\frac{(x-1)^2}{\left(\sqrt2+\sqrt{3-x}\right)^2}\geq0.$$
Also, by Jensen we obtain:
$$\sum_{cyc}\sqrt{x+y}\leq3\sqrt{\frac{\sum\limits_{cyc}(x+y)}{3}}=\sqrt{6(x+y+z)}.$$
