Prove that $\frac{x^2y}{z}+\frac{y^2z}{x}+\frac{z^2x}{y} \geq x^2+y^2+z^2$

Question

Let $x,y,z \in \mathbb{R}$ such that $x \geq y \geq z > 0$. Prove that $$\frac{x^2y}{z}+\frac{y^2z}{x}+\frac{z^2x}{y} \geq x^2+y^2+z^2.$$

I rearranged the inequality to get $$xy(x^2y)+yz(y^2z)+xz(z^2x) \geq xyz(x^2+y^2+z^2).$$ I then thought about using the rearrangement inequality but didn't see how to use it. How can we continue?

Answer 1

We need to prove that $\sum\limits_{cyc}(x^3y^2-x^3yz)\geq0$ or

$$\sum\limits_{cyc}(z^3x^2+z^3y^2-2z^3xy)\geq\sum\limits_{cyc}(x^3z^2-x^3y^2)$$ or $$\sum\limits_{cyc}z^3(x-y)^2\geq(xy+xz+yz)(x-y)(y-z)(z-x)$$

which is obvious.