Prove that if $x > 3$ and $y < 2$, then $x^2 − 2y > 5$ 
Suppose $x > 3$ and $y < 2$. Then $x^2 − 2y > 5$.

My attempt: 


*

*$x>3 \implies x^2 > 9$

*$ y < 2 \implies 5 + 2y < 9$
It follows that $5+2y < 9 < x^2$. From this we can see that  $5+2y < x^2 \implies x^2 -2y > 5$ 
Therefore, if $x>3$ and $y<2$, then $x^2 − 2y > 5$.
Is it accurate?

Yes, your proof is correct. Here is an alternative:
$x^2> 9$ and since $-y>-2$ we have $-2y>-4$, so $$x^2-2y>9-4 =5$$
