Let $a, b \in \mathbb{R}$. Show that $|ab| \le \frac 12 (a^2 + b^2 )$.
Need consider three cases:
(i) both $a,b \ge 0$: $|ab| = ab=\frac{2ab}2= \frac{(a+b)^2-a^2-b^2}2=\frac{(a+b)^2-(a^2+b^2)}2$.
(ii) both $a,b \lt 0$: $|ab| = ab$ , same as in (i).
(iii) either one of $a,b \lt 0$: $|ab| = -ab=-\frac{2ab}2= -(\frac{(a+b)^2-a^2-b^2}2)=-(\frac{(a+b)^2-(a^2+b^2)}2)$.
So, need prove $2ab \le a^2+b^2$ which is again back to the question.
I have in mind only an approach based on triangle formed by three vectors, but am unclear about using that. The reason is need to constrain the angle formed to $90^o$, as only then $a^2 + b^2 = a^2+b^2$.
But, even with that constraint, the issue is that the vector approach has equivalent formulation only in terms of complex quantities.
So, no progress possible.