Prove That $ \sqrt{\frac xy + \frac yz}+ \sqrt{\frac yz + \frac zx}+ \sqrt{\frac zx + \frac xy}\ge 3 $

Question

for x, y, z $\gt 0$, prove that: $$ \sqrt{\frac xy + \frac yz}+ \sqrt{\frac yz + \frac zx}+ \sqrt{\frac zx + \frac xy}\ge 3 $$

Answer 1

By AM-GM inequality we have $$ \begin{align*} \frac{x}{y}+\frac{y}{z} &\ge 2\sqrt{\frac{x}{z}}\\ \frac{y}{z}+\frac{z}{x} &\ge 2\sqrt{\frac{y}{x}}\\ \frac{z}{x}+\frac{x}{y} &\ge 2\sqrt{\frac{x}{z}}. \end{align*} $$ Then again by AM-GM inequality we have $$ \begin{align*} \frac{1}{3}\left(\sqrt{\frac{x}{y}+\frac{y}{z}}+\sqrt{\frac{y}{z}+\frac{z}{x}}+\sqrt{\frac{z}{x}+\frac{x}{y}} \right) &\ge \sqrt[3]{\sqrt{\frac{x}{y}+\frac{y}{z}}\sqrt{\frac{y}{z}+\frac{z}{x}}\sqrt{\frac{z}{x}+\frac{x}{y}}}\\ &\ge \sqrt[3]{\sqrt{2\sqrt{\frac{x}{z}}}\sqrt{2\sqrt{\frac{y}{x}}}\sqrt{2\sqrt{\frac{z}{y}}}}\\ &=2^{1/2}\left(\frac{xyz}{zxy}\right)^{1/12}\\ &> 1, \end{align*} $$ and our result follows.

Answer 2

Also, we can use Minkowski and AM-GM $$\sum_{cyc}\sqrt{\frac{x}{y}+\frac{y}{z}}\geq\sqrt{\left(\sum_{cyc}\sqrt{\frac{x}{y}}\right)^2+\left(\sum_{cyc}\sqrt{\frac{y}{z}}\right)^2}\geq\sqrt{3^2+3^2}=3\sqrt2\geq3.$$