Function of modulus value

Question

If $f(x)=|x+2|$ and $g(x)=2|x|$, then it is true that $\frac{f(x)+g(x)-|f(x)-g(x)|}{2}=Min(f(x),g(x))$, How do we prove it.

Answer 1

This expression is always true, you can check that :

if $f\leq g$, then : $$ \frac{f+g-|f-g|}{2}=\frac{f+g-(g-f)}{2}=f$$ Then, if $g\leq$ f : $$\frac{f+g-|f-g|}{2}=\frac{f+g-(f-g)}{2}=g $$

So you get that : $$\frac{f+g-|f-g|}{2}=\min(f,g) $$