<s>[INST] Prove that $\sqrt{x^2+yz}+\sqrt{y^2+xz}+\sqrt{z^2+xy} \le 3$ Prove that 
$$\sqrt{x^2+yz}+\sqrt{y^2+xz}+\sqrt{z^2+xy} \le 3$$
for $x \ge 0$, $y \ge 0$, $z \ge 0$ and $x+y+z \le 2$.
My work: 
\begin{align*}
&\mathrel{\phantom{=}} \sqrt{x^2+yz}+\sqrt{y^2+xz}+\sqrt{z^2+xy}\\
&\le\sqrt3 \sqrt{x^2+yz+y^2+xz+z^2+xy}\\
&\le\sqrt3\sqrt{2x^2+2y^2+2z^2}=\sqrt6\sqrt{x^2+y^2+z^2}.
\end{align*}
 [/INST] It is enough to show the homogeneous inequality $\sum_{cyc} \sqrt{x^2+yz} \leqslant \frac32(x+y+z)$.  WLOG, let $x \geqslant y \geqslant z$.  
Note by AM-GM, we have $\sqrt{x^2+yz} \leqslant \sqrt{x^2+xz} \leqslant x + \frac12z$.  This type of AM-GM is motivated by noting $(1, 1, 0)$ is a solution for equality.
Further by CS (or power means), we get
$$\sqrt{y^2+zx} + \sqrt{z^2+xy} \leqslant \sqrt{1+1} \sqrt{(y^2+zx)+(z^2+xy)}$$
Thus it is enough to show that
$$\sqrt2\sqrt{y^2+z^2+x(y+z)} \leqslant \frac12(x+3y+2z)$$
Simplifying, this is $(x-y-2z)^2+8z(y-z) \geqslant 0$ which is obvious.  Equality is when any two variables are equal and the third zero.
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