Prove $x^2+y^2< 1$ (Continued)

On this blog post, I will still discuss the problem how to prove $x^2+y^2< 1$ given $x,\,y>0$ and $x^3+y^3<x-y$.

Previously, I have mentioned about how important it is for us to look at what are given, assimilate it next  and jot down if we are given something that is/are implicitly correct.

It's utterly important to recognize in this intriguing problem that $x-y>0$ since both $x,\,y>0$ and hence $x^3+y^3>0$. This suggests $x-y>0$ since $x-y>x^3+y^3>0$.

Back to what we are given, if we start with what have, i.e.

$x-y>x^3+y^3$

We see that $x^3+y^3>x^3-y^3$ and so we obtain

$x-y>x^3-y^3-(*)$

But we know fairly well that $x^3-y^3$ can be factored perfectly where $x^3-y^3=(x-y)(x^2+xy+y^2)$

Replace the factored form of the difference of cubes to the inequality (*), we see that

$x-y>(x-y)(x^2+xy+y^2)-(**)$

The only issue that we have now is, if we could divide this inequality by the quantity of $x-y$, if it is a negative, then we need to revert the inequality sign but if it is a positive quantity, then we can safely do the division without changing the inequality sign, that is why I stress the importance of recognizing something implicitly true from the given information.

Since we have $x-y>0$ for all $x,\,y>0$, we can divide the inequality (**) by $x-y$ and this yields

$\cancel{x-y}>\cancel{(x-y)}(x^2+xy+y^2)$

$1>x^2+xy+y^2>x^2+y^2$ (Q.E.D.)