Find $\frac{x^2}{y} + \frac{y^2}{z} + \frac{z^2}{x} \to \min$ I'm trying to solve
\begin{align*}
&\frac{x^2}{y} + \frac{y^2}{z} + \frac{z^2}{x} \to \min \\
&x + y + z = 1 \\
&x, y, z > 0,
\end{align*}
using only inequalities.
How can i solve it? I used am-gm, tried to "break" terms, but got only after all $(xy + yz + xz) = \frac{1}{2}$

By Cauchy-Schwarz we get$$\frac{x^2}{y} + \frac{y^2}{z} + \frac{z^2}{x} \geq \frac{(x+y+z)^2}{x+y+z}=1$$
