Prove that $\sqrt{x^2+yz}+\sqrt{y^2+xz}+\sqrt{z^2+xy} \le 3$

Question

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*}

Answer 1

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.

Answer 2

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: $$2\sum_{cyc}(x^8-x^7y-x^7z+x^6yz)\geq0.$$