Problem 1604

Prove that by Cauchy-Schwarz inequality,
$$\begin{array}{l} \left(\frac{x^{2}}{\sqrt{2 x^{2}(y+z)}}+\frac{y^{2}}{\sqrt{2 y^{2}(z+x)}}+\frac{z^{2}}{\sqrt{2 z^{2}(x+y)}}\right)(\sqrt{2(y+z)}+\sqrt{2(z+x)}+\sqrt{2(x+y)}) \geqslant \\ (\sqrt{x}+\sqrt{y}+\sqrt{z})^{2}=1 \\ \left(\frac{y z}{\sqrt{2 x^{2}(y+z)}}+\frac{z x}{\sqrt{2 y^{2}(z+x)}}+\frac{x y}{\sqrt{2 z^{2}(x+y)}}\right)(\sqrt{2(y+z)}+\sqrt{2(z+x)}+\sqrt{2(x+y)}) \geqslant \\ \left(\sqrt{\frac{y z}{x}}+\sqrt{\frac{z x}{y}}+\sqrt{\frac{x y}{z}}\right)^{2} \end{array}$$
Adding (1) and (2) and using the AM-GM inequality, we get
$$\begin{array}{l} \left(\frac{x^{2}+y z}{\sqrt{2 x^{2}(y+z)}}+\frac{y^{2}+z x}{\sqrt{2 y^{2}(z+x)}}+\frac{z^{2}+x y}{\sqrt{2 z^{2}(x+y)}}\right)(\sqrt{2(y+z)}+\sqrt{2(z+x)}+\sqrt{2(x+y)}) \geqslant \\ 1+\left(\sqrt{\frac{y z}{x}}+\sqrt{\frac{z x}{y}}+\sqrt{\frac{x y}{z}}\right)^{2} \geqslant 2\left(\sqrt{\frac{y z}{x}}+\sqrt{\frac{z x}{y}}+\sqrt{\frac{x y}{z}}\right) \end{array}$$

Now, we need to prove
$$2\left(\sqrt{\frac{y z}{x}}+\sqrt{\frac{z x}{y}}+\sqrt{\frac{x y}{z}}\right) \geqslant \sqrt{2(y+z)}+\sqrt{2(z+x)}+\sqrt{2(x+y)}$$

By the AM-GM inequality,
$$\left[\sqrt{\frac{y z}{x}}+\left(\frac{1}{2} \sqrt{\frac{z x}{y}}+\frac{1}{2} \sqrt{\frac{x y}{z}}\right)\right]^{2} \geqslant 4 \sqrt{\frac{y z}{x}}\left(\frac{1}{2} \sqrt{\frac{z x}{y}}+\frac{1}{2} \sqrt{\frac{x y}{z}}\right)=2(y+z)$$

Thus,
$$\sqrt{\frac{y z}{x}}+\left(\frac{1}{2} \sqrt{\frac{z x}{y}}+\frac{1}{2} \sqrt{\frac{x y}{z}}\right) \geqslant \sqrt{2(y+z)}$$

Similarly,
$$\begin{array}{l} \sqrt{\frac{z x}{y}}+\left(\frac{1}{2} \sqrt{\frac{x y}{z}}+\frac{1}{2} \sqrt{\frac{y z}{x}}\right) \geqslant \sqrt{2(z+x)} \\ \sqrt{\frac{x y}{z}}+\left(\frac{1}{2} \sqrt{\frac{y z}{x}}+\frac{1}{2} \sqrt{\frac{z x}{y}}\right) \geqslant \sqrt{2(x+y)} \end{array}$$
Adding (4), (5), and (6) gives inequality (3), thus proving the original inequality.