Problem - 3228
Let $x, y, z$ be strictly positive real numbers. Prove that $$\Big(\frac{x}{y}+\frac{z}{\sqrt[3]{xyz}}\Big)^2+\Big(\frac{y}{z}+\frac{x}{\sqrt[3]{xyz}}\Big)^2+\Big(\frac{z}{x}+\frac{y}{\sqrt[3]{xyz}}\Big)^2 \ge 12$$
By AM-QM:
$$\frac{1}{3}\Big[\Big(\frac{x}{y}+\frac{z}{\sqrt[3]{xyz}}\Big)^2+\Big(\frac{y}{z}+\frac{x}{\sqrt[3]{xyz}}\Big)^2+\Big(\frac{z}{x}+\frac{y}{\sqrt[3]{xyz}}\Big)^2\Big]\ge \frac{1}{9}\Big(\frac{x}{y}+\frac{y}{z}+\frac{z}{x}+\frac{x+y+z}{\sqrt[3]{xyz}}\Big)^2$$
Therefore,
\begin{align*}
&\Big(\frac{x}{y}+\frac{z}{\sqrt[3]{xyz}}\Big)^2+\Big(\frac{y}{z}+\frac{x}{\sqrt[3]{xyz}}\Big)^2+\Big(\frac{z}{x}+\frac{y}{\sqrt[3]{xyz}}\Big)^2\\
\ge\ &\frac{1}{3}\Big(\frac{x}{y}+\frac{y}{z}+\frac{z}{x}+\frac{x+y+z}{\sqrt[3]{xyz}}\Big)^2\\
\ge\ &\frac{1}{3}\Big(3\sqrt[3]{\frac{x}{y}\cdot\frac{y}{z}\cdot\frac{z}{x}}+\frac{x+y+z}{\sqrt[3]{xyz}}\Big)^2\\
=\ &\frac{1}{3}\Big(3+\frac{x+y+z}{\sqrt[3]{xyz}}\Big)\\
=\ &\ge12
\end{align*}