Inequality

[requferofile]

if$x >y>0$
prove that
$y(x-y)(x-3)\geq -1$

[Руст]

If $x\ge 3$ left side is positive.
If $x<3$, then $x(x-y)\le \frac{x^2}{4}$, therefore
$x(x-y)(x-3)\ge f(x)=\frac{x^3-3x^2}{4}$. Obviosly minimal value $f(x)=f(2)=-1.$