If $x,y,z \in \mathbb{R}^+$, then prove that $\frac{x^2}{y} + \frac{y^2}{z} + \frac{z^2}{x}\geq x+y+z$.

Question

If $x,y,z \in \mathbb{R}^+$, then prove that $$\frac{x^2}{y} + \frac{y^2}{z} + \frac{z^2}{x}\geq x+y+z\,.$$

What are other proofs? Here is my solution.

Note by the AM-GM Inequality that $$\frac{a^2}{b}+b\geq 2\sqrt{\left(\frac{a^2}{b}\right)\cdot b}=2a$$ for all $a,b>0$. Therefore, $$\left(\frac{x^2}{y}+y\right)+\left(\frac{y^2}{z}+z\right)+\left(\frac{z^2}{x}+x\right)\geq 2x+2y+2z\,.$$ This means $$\left(\frac{x^2}{y}+\frac{y^2}{z}+\frac{z^2}{x}\right)+(x+y+z)\geq 2(x+y+z)\,,$$ which is equivalent to the inequality to be proven.

Answer 1

By C-S $$\sum_{cyc}\frac{x^2}{y}\geq\frac{(x+y+z)^2}{\sum\limits_{cyc}y}=x+y+z.$$ Also, we have $$\sum_{cyc}\frac{x^2}{y}=\sum_{cyc}\left(\frac{x^2}{y}-2x+y\right)+x+y+z=\sum_{cyc}\frac{(x-y)^2}{y}+x+y+z\geq x+y+z.$$ The last solution we can write by AM-GM: $$\sum_{cyc}\frac{x^2}{y}=\sum_{cyc}\left(\frac{x^2}{y}+y\right)-x-y-z\geq\sum_{cyc}2\sqrt{\frac{x^2}{y}\cdot y}-x-y-z=x+y+z.$$ Actually, the following much more stronger inequality is also true.

Let $x$, $y$ and $z$ be positives such that $x^6+y^6+z^6=3.$ Prove that:

$$\frac{x^2}{y}+\frac{y^2}{z}+\frac{z^2}{x}\geq3.$$