Find $\frac{x^2}{y} + \frac{y^2}{z} + \frac{z^2}{x} \to \min$

Question

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 $(xy + yz + xz) = \frac{1}{2}$

Answer 1

You can use Cauchy-Schwarz: $$\sum\dfrac{x^2}{y}\sum y\geq(x+y+z)^2$$ or an even simpler one is: $$\sum\dfrac{x^2}{y}-\sum y = \sum\left(\dfrac{x^2}{y} - 2x+y\right)=\sum\dfrac{(x-y)^2}{y}\geq 0.$$

Answer 2

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