<s>[INST] If $a,b \in \Bbb R$, prove that $|ab| \le (a^2+b^2)/2$ So far I have the first case when $a=b$:
\begin{align*}
|ab|
&= |b^2|\\
&=|b|^2\\
&=\frac{2|b|^2}2\\
&=\frac{b^2+b^2}2\\
&=\frac{a^2+b^2}2
\end{align*}
Case 2: $a>b$
Case 3 $a<b$
I've been stuck on this problem for a few hours now and don't know how to proceed. Am I approaching it wrong? Should I not be thinking about the cases of $a$ in terms of $b$? Thank you for your help.
 [/INST] If $ab\ge0$ then $|ab|=ab\le ab+\dfrac{(a-b)^2}2=\dfrac{a^2+b^2}2$.
If $ab\le0$ then $|ab|=-ab\le-ab+\dfrac{(a+b)^2}2=\dfrac{a^2+b^2}2$.
</s>