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

Question

Let $x,y,z$ be lengths of three sides of triangle. Show that $\frac{x^2y}{z}+\frac{y^2z}{x}+\frac{xz^2}{y}\geq x^2+y^2+z^2$. This seems simple but not easy. I try to use RAVI replacement but it doesn't work. I think it has a relation to the inequality $x^2y(x-y)+y^2z(y-z)+z^2x(z-x)$ ( which appears in $IMO$ $1983$.). Give me some idea to solve this, thank you so much!

Answer 1

The hint.

Use the BW:

Let $x=a+u$, $y=a+v$ and $z=a+u+v$, where $a>0$, $u\geq0$ and $v\geq0.$

After this substitution you'll get something obvious.