Prove if $x > 3$ and $y < 2$, then $x^{2} - 2y > 5$ My solution is:
Multiply $x > 3$ with $x$, yielding $x^{2} > 9$
Multiply $y < 2$ with $2$, yielding $2y < 4$
Thus, based on the above $2$ yielded inequalities, we can prove that if $x > 3$ and $y < 2$, then $x^{2} - 2y > 5$.
Is this a correct proofing steps?

For a roundabout way to prove it, which is overkill in this case, but may prove useful in other cases, note that the blue terms are positive since $x -3\gt 0$ and $2-y \gt 0\,$, therefore:
$$
x^2-2y =\left((x-3)+3\right)^2 - 2\left(-(2-y)+2\right) = \color{blue}{(x-3)^2} + 6\color{blue}{(x-3)} + \color{red}{9} +2\color{blue}{(2-y)} - \color{red}{4} \gt \color{red}{5}
$$
