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?
 A: Just $x^2-2y>3^2-2\cdot2=5$
A: Looks good!
$$x>3 \implies x^2>9$$
$$y<2 \implies 2y<4 \implies -2y>-4$$
Add the two to get: $$x^2-2y>5$$
A: 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:
$$
x^2-2y =\left((x-3)+3\right)^2 - 2\left(-(2-y)+2\right) = \color{blue}{(x-3)^2} + 6\color{blue}{(x-3)} + \color{red}{9} +2\color{blue}{(2-y)} - \color{red}{4} \gt \color{red}{5}
$$
A: $x > 3,$ $y < 2$ implies $(-1)y > -2$ which is equal to $(-2)y > (-4).$ Hence we get $x^2 - 2y > 9 - 4 = 5.$
A: As a slightly extended version of Michael Rozenberg's answer, this can very simply be written down as:$%
\newcommand{\calc}{\begin{align} \quad &}
\newcommand{\op}[1]{\\ #1 \quad & \quad \unicode{x201c}}
\newcommand{\hints}[1]{\mbox{#1} \\ \quad & \quad \phantom{\unicode{x201c}} }
\newcommand{\hint}[1]{\mbox{#1} \unicode{x201d} \\ \quad & }
\newcommand{\endcalc}{\end{align}}
\newcommand{\subcalch}[1]{\\ \quad & \quad #1 \\ \quad &}
\newcommand{\subcalc}{\quad \begin{aligned} \quad & \\ \bullet \quad & }
\newcommand{\endsubcalc}{\end{aligned} \\ \\ \cdot \quad &}
\newcommand{\ref}[1]{\text{(#1)}}
\newcommand{\then}{\Rightarrow}
\newcommand{\when}{\Leftarrow}
%$
$$\calc
    x^2-2y
\op>\hints{using $\;x > 3\;$ and the fact that $\;a-b\;$ is monotonic in $\;a\;$;}
    \hint{using $\;y < 2\;$ and the fact that $\;a-b\;$ is antimonotonic in $\;b\;$}
    3^2-2\times2
\op=\hint{arithmetic}
    5
\endcalc$$
In my opinion, this proof is the most direct reflection of the basic idea of this proof.
