How to prove $\frac{1}{2}(x^2+y^2) \geq |xy|$

Question

I understand how to get $\dfrac{1}{2}(x^2+y^2) \geq xy$ starting from $(x-y)^2 \geq 0$, but I don't understand how to get the absolute value version.

Answer 1

Just replace $x,y$ by their absolute values and it’s still true in your argument and notice that : $|x||y| = |xy|$.