Lin Algebra Simple Proof of identity

Question

I kind of have an idea how to do it but don't know how to follow through with proving:

Prove:

$|ab| \le \frac 12(a^2+b^2)$

This is what I have so far, not sure how to complete the proof:

Pf.

$2|ab| \le (a^2+b^2)$

$0\le a^2+b^2-2|ab| $

Then I think it has something to do with $a^2$ and $b^2$ are always positive.

Answer 1

By using of your work we obtain: $$a^2+b^2-2|ab|=|a|^2|+|b|^2-2|a||b|=(|a|-|b|)^2\geq0.$$ Done!

Answer 2

Hint: Not exactly, but the idea is to use squares, yes. If you play with $(a+b)^2 \geq 0$ you'll get $a^2 + b^2 \geq -2ab$, and so $(a^2+b^2)/2 \geq -ab$. If you start with $(a-b)^2 \geq 0$ instead you'll get $(a^2+b^2)/2 \geq ab$. Together, these two inequalities mean that $(a^2+b^2)/2 \geq |ab|$, as wanted. I'll let you fill the details, it is instructive.