Trying to check my answer for this question:
Let $f,g$ be continous real functions in closed interval $[a,b]$, such that $\forall x\in [a,b], f(x)>g(x)$. Then $\sup f([a,b])>\sup g([a,b])$.
It seems obviously true - unless someone would come up with a surprising counter-example I have yet to seen.
This is my proof: Since $f,g$ are both continous in $[a,b]$, according to the extreme value theorem, $\exists c\in[a,b], f(c)=\max f([a,b])=\sup f([a,b])$ , and also $\exists d\in[a,b], g(d)=\max g([a,b])=\sup g([a,b])$.
So, ultimatelly: $f(c)\geq f(d)>g(d)$ and it's setlled. Right?
I could also use the defenition of the supremum with $\forall \epsilon>0 \exists x\in[a,b]$ such that $ f(x)>\sup f([a,b])-\epsilon$ after showing its existence with the EVT, but it all goes down to that same inequality.
am I right?