Proving inequality $(x^2+y^2)(y-1)+yx-y^2<0$

Question

I have an inequality which came out of Lyapunov function for system of ODE's: $$(x^2+y^2)(y-1)+yx-y^2<0.$$ To prove stability of my solution, I have to prove that the inequalty is true in area $0<x^2 +y^2<1$. I know it must be simple, but I am still in trouble. Any help is appreciated.

Answer 1

If $x = y$, then $$(x^2 + y^2)(y - 1) + yx - y^2 = 2x^2(x - 1) < 0$$ since $x < 1$. Now suppose $x \neq y$. If $x$ and $y$ are positive, then $0 < x^2 < x$ and $y^3 < y^2$, thus

\begin{align}(x^2 + y^2)(y - 1) + yx - y^2 &= x^2y + y^3 - x^2 - y^2 + yx - y^2\\ & < x^2y - x^2 - y^2 + yx \\ & < 2xy - (x^2 + y^2)\\ & < 0.\end{align}

If $x < 0 < y$ or $y < 0 < x$, then $yx - y^2 = y(x - y) < 0$; since $(x^2 + y^2)(y - 1) < 0$ (as $y < 1$) we have $(x^2 + y^2)(y - 1) + yx - y^2 < 0$. Finally, if $x$ and $y$ are both negative, then $y^3 < 0 < y^2$ and $x^2 > 0 > x$; like in the case when $x$ and $y$ are both positive, we have

$$(x^2 + y^2)(y - 1) + yx - y^2 < x^2y - x^2 - y^2 + yx < 2xy - (x^2 + y^2) < 0.$$