XIX Asian Pacific Mathematics Olympiad

We first note that $$\begin{array}{rl}\frac{x^2+yz}{\sqrt{2x^2(y+z)}}& =\frac{x^2-x(y+z)+yz}{\sqrt{2x^2(y+z)}}+\frac{x(y+z)}{\sqrt{2x^2(y+z)}}\\ & =\frac{(x-y)(x-z)}{\sqrt{2x^2(y+z)}}+\sqrt{\frac{y+z}{2}}\\ & \ge \frac{(x-y)(x-z)}{\sqrt{2x^2(y+z)}}+\frac{\sqrt{y}+\sqrt{z}}{2}\end{array}\text{\hspace{1em}(1)}$$ Similarly, we have $$\begin{array}{rl}& \frac{y^2+zx}{\sqrt{2y^2(z+x)}}\ge \frac{(y-z)(y-x)}{\sqrt{2y^2(z+x)}}+\frac{\sqrt{z}+\sqrt{x}}{2}\\ & \frac{z^2+xy}{\sqrt{2z^2(x+y)}}\ge \frac{(z-x)(z-y)}{\sqrt{2z^2(x+y)}}+\frac{\sqrt{x}+\sqrt{y}}{2}\end{array}\text{\hspace{1em}(2)(3)}$$ We now add (1)~(3) to get $$\begin{array}{rl}& \frac{x^2+yz}{\sqrt{2x^2(y+z)}}+\frac{y^2+zx}{\sqrt{2y^2(z+x)}}+\frac{z^2+xy}{\sqrt{2z^2(x+y)}}\\ & \ge \frac{(x-y)(x-z)}{\sqrt{2x^2(y+z)}}+\frac{(y-z)(y-x)}{\sqrt{2y^2(z+x)}}+\frac{(z-x)(z-y)}{\sqrt{2z^2(x+y)}}+\sqrt{x}+\sqrt{y}+\sqrt{z}\\ & =\frac{(x-y)(x-z)}{\sqrt{2x^2(y+z)}}+\frac{(y-z)(y-x)}{\sqrt{2y^2(z+x)}}+\frac{(z-x)(z-y)}{\sqrt{2z^2(x+y)}}+1\end{array}$$ Thus, it suffices to show that $$\begin{array}{c}\frac{(x-y)(x-z)}{\sqrt{2x^2(y+z)}}+\frac{(y-z)(y-x)}{\sqrt{2y^2(z+x)}}+\frac{(z-x)(z-y)}{\sqrt{2z^2(x+y)}}\ge 0\end{array}\text{\hspace{1em}(4)}$$ Now, assume without loss of generality, that $x\ge y\ge z$. Then we have $$\frac{(x-y)(x-z)}{\sqrt{2x^2(y+z)}}\ge 0$$ and $$\begin{array}{rl}& \frac{(z-x)(z-y)}{\sqrt{2z^2(x+y)}}+\frac{(y-z)(y-x)}{\sqrt{2y^2(z+x)}}=\frac{(y-z)(x-z)}{\sqrt{2z^2(x+y)}}-\frac{(y-z)(x-y)}{\sqrt{2y^2(z+x)}}\\ & \ge \frac{(y-z)(x-y)}{\sqrt{2z^2(x+y)}}-\frac{(y-z)(x-y)}{\sqrt{2y^2(z+x)}}=(y-z)(x-y)\left(\frac{1}{\sqrt{2z^2(x+y)}}-\frac{1}{\sqrt{2y^2(z+x)}}\right).\end{array}$$ The last quantity is non-negative due to the fact that $$y^2(z+x)=y^{2}z+y^{2}x\ge y{z}^2+z^{2}x=z^2(x+y).$$ This completes the proof.

Second solution:

By Cauchy-Schwarz inequality, $$\begin{array}{rl}& \left(\frac{x^2}{\sqrt{2x^2(y+z)}}+\frac{y^2}{\sqrt{2y^2(z+x)}}+\frac{z^2}{\sqrt{2z^2(x+y)}}\right)\\ & \times (\sqrt{2(y+z)}+\sqrt{2(z+x)}+\sqrt{2(x+y)})\ge (\sqrt{x}+\sqrt{y}+\sqrt{z}{)}^2=1,\end{array}\text{\hspace{1em}(5)}$$ and $$\begin{array}{rl}& \left(\frac{yz}{\sqrt{2x^2(y+z)}}+\frac{zx}{\sqrt{2y^2(z+x)}}+\frac{xy}{\sqrt{2z^2(x+y)}}\right)\\ & \times (\sqrt{2(y+z)}+\sqrt{2(z+x)}+\sqrt{2(x+y)})\ge {\left(\sqrt{\frac{yz}{x}}+\sqrt{\frac{zx}{y}}+\sqrt{\frac{xy}{z}}\right)}^2.\end{array}\text{\hspace{1em}(6)}$$ We now combine (5) and (6) to find $$\begin{array}{rl}& \left(\frac{x^2+yz}{\sqrt{2x^2(y+z)}}+\frac{y^2+zx}{\sqrt{2y^2(z+x)}}+\frac{z^2+xy}{\sqrt{2z^2(x+y)}}\right)\\ & \times (\sqrt{2(x+y)}+\sqrt{2(y+z)}+\sqrt{2(z+x)})\\ & \ge 1+{\left(\sqrt{\frac{yz}{x}}+\sqrt{\frac{zx}{y}}+\sqrt{\frac{xy}{z}}\right)}^2\ge 2\left(\sqrt{\frac{yz}{x}}+\sqrt{\frac{zx}{y}}+\sqrt{\frac{xy}{z}}\right).\end{array}$$ Thus, it suffices to show that $$\begin{array}{c}2\left(\sqrt{\frac{yz}{x}}+\sqrt{\frac{zx}{y}}+\sqrt{\frac{xy}{z}}\right)\ge \sqrt{2(y+z)}+\sqrt{2(z+x)}+\sqrt{2(x+y)}.\end{array}\text{\hspace{1em}(7)}$$ Consider the following inequality using AM-GM inequality $${\left[\sqrt{\frac{yz}{x}}+\left(\frac{1}{2}\sqrt{\frac{zx}{y}}+\frac{1}{2}\sqrt{\frac{xy}{z}}\right)\right]}^2\ge 4\sqrt{\frac{yz}{x}}\left(\frac{1}{2}\sqrt{\frac{zx}{y}}+\frac{1}{2}\sqrt{\frac{xy}{z}}\right)=2(y+z),$$ or equivalently $$\sqrt{\frac{yz}{x}}+\left(\frac{1}{2}\sqrt{\frac{zx}{y}}+\frac{1}{2}\sqrt{\frac{xy}{z}}\right)\ge \sqrt{2(y+z)}$$ Similarly, we have $$\begin{array}{rl}& \sqrt{\frac{zx}{y}}+\left(\frac{1}{2}\sqrt{\frac{xy}{z}}+\frac{1}{2}\sqrt{\frac{yz}{x}}\right)\ge \sqrt{2(z+x)}\\ & \sqrt{\frac{xy}{z}}+\left(\frac{1}{2}\sqrt{\frac{yz}{x}}+\frac{1}{2}\sqrt{\frac{zx}{y}}\right)\ge \sqrt{2(x+y)}\end{array}$$ Adding the last three inequalities, we get $$2\left(\sqrt{\frac{yz}{x}}+\sqrt{\frac{zx}{y}}+\sqrt{\frac{xy}{z}}\right)\ge \sqrt{2(y+z)}+\sqrt{2(z+x)}+\sqrt{2(x+y)}.$$ This completes the proof.