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

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.
