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

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!

• The inequality is not rotational. Are you sure the question is correct? – Hw Chu Jul 24 at 13:02
• @HwChu oh i am sorry so much. I have already change my mistake. – user628755 Jul 24 at 13:12

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