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*}
 A: 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.
A: By C-S we have:
$$\left(\sum_{cyc}\sqrt{x^2+yz}\right)^2\leq\sum_{cyc}\frac{x^2+yz}{2x^2+y^2+z^2+yz}\sum_{cyc}(2x^2+y^2+z^2+yz).$$
Id est, it remains to prove that
$$\sum_{cyc}\frac{x^2+yz}{2x^2+y^2+z^2+yz}\sum_{cyc}(4x^2+yz)\leq\frac{9}{4}(x+y+z)^2,$$ which is
$$\sum_{sym}(x^8+2x^7y+9x^6y^2-23x^5y^3+10x^4y^4)+$$
$$+\sum_{sym}(5.5x^6yz+70x^5y^2z+79x^4y^3z+131.5x^4y^2z^2-97.5x^3y^3z^2)\geq0,$$
which is obvious after using following Schur:
$$2\sum_{cyc}(x^8-x^7y-x^7z+x^6yz)\geq0.$$
