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?

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.
