Does $-\min(a,b) = \max(-a,-b)$?

Solution 1:

It's true, yes. And it's a mere consequence of $$a\leq b\iff -a\geq-b$$ so that if this is true then $$-\min(a,b)=-a = \max(-a,-b),$$ and a symmetric argument solves the case $b\leq a$

Solution 2:

Alternatively, use $\min(x,\,y)=\frac{x+y-|x-y|}{2},\,\max(x,\,y)=\frac{x+y+|x-y|}{2}$, so the first equation reduces to $-\frac{a+b-|a-b|}{2}=\frac{-a-b+|-a+b|}{2}$, which is trivial. The second equation follows similarly.

Or even simpler, since both claims are symmetric in $a,\,b$, without loss of generality $a\le b$ so $-b\le a$. Then the claims respectively reduce to $-a=-a,\,-b=-b$.