Find $\frac{x^2}{y} + \frac{y^2}{z} + \frac{z^2}{x} \to \min$ I'm trying to solve
\begin{align*}
&amp;\frac{x^2}{y} + \frac{y^2}{z} + \frac{z^2}{x} \to \min \\
&amp;x + y + z = 1 \\
&amp;x, y, z &gt; 0,
\end{align*}
using only inequalities.
How can i solve it? I used am-gm, tried to "break" terms, but got only after all $(xy + yz + xz) = \frac{1}{2}$
You can use Cauchy-Schwarz:
$$\sum\dfrac{x^2}{y}\sum y\geq(x+y+z)^2$$
or an even simpler one is:
$$\sum\dfrac{x^2}{y}-\sum y = \sum\left(\dfrac{x^2}{y} - 2x+y\right)=\sum\dfrac{(x-y)^2}{y}\geq 0.$$
By Cauchy-Schwarz we get$$\frac{x^2}{y} + \frac{y^2}{z} + \frac{z^2}{x} \geq \frac{(x+y+z)^2}{x+y+z}=1$$