<s>[INST] How to divide two inequalities I would like to know if somebody knows how to properly divide one inequality by another, as a resolution method similar to when we divide one equality by another.
Take this as an example: $x^2 - y^2 \lt 8$ and $x + y \gt 3$.
I want to know if it is possible to factor the left hand side of the first equation, to multiply by $-1$ the second equation and divide both equation as to obtain: $y - x \gt -8/3$. I don't really know if the sign must be < or >.
Thanks!
 [/INST] Suppose $x^2-y^2 < 8$ and $x+y > 3$. If $y - x \le -\dfrac{8}{3}$, then $x-y \ge \dfrac{8}{3}$, and hence $x^2-y^2 = (x-y)(x+y) > \dfrac{8}{3} \cdot 3 = 8$, a contradiction since $x^2-y^2 < 8$. 
Thus, any $(x,y)$ which satisfy both $x^2-y^2 < 8$ and $x+y > 3$ will also satisfy $y - x < -\dfrac{8}{3}$. 
However, the converse is not true, that is if $(x,y)$ satisfy $y-x < -\dfrac{8}{3}$, then it is not necessarily true that $x^2-y^2 < 8$ and $x+y > 3$. See the other comments for counterexamples. 
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