Question: Question:
Prove that if $f, g: [a, b] → \mathbb{R}$ are continuous and $f (x) <g (x) $ for every $x \in [a, b]$, then there exists $δ> 0$ such that $f + δ ≤ g$.

I have tried to take the $f-g$ function and apply something like intermediate value theorem, but couldn't conclude anything. I've seen a [solution](https://math.stackexchange.com/questions/379843/prove-for-continuous-f-and-g-fxgx-there-exists-k-such-that-fxk) that uses covers, but we haven't seen that yet and would like to know if there is an easier way to fix it.

Answer: 
let $h=g-f>0$ on $[a,b]$. Since $h$ is continuous on a closed interval, it has a global minimum (extreme value theorem).

then let $\delta=\min\_{x\in[a,b]}{h(x)}$

Then, by definition $\delta>0$ and $\forall x\in[a,b], f(x)+\delta\leq g(x) $
Suppose not. Then for every $\delta>0$ there is a point $x\in[a,b]$ where $g(x)-f(x)<\delta$. In particular, there is a sequence $x\_n\in[a,b]$ with $g(x\_n)-f(x\_n)<1/n$.

The sequence $x\_n$ doesn't necessarily have a limit, but by Bolzano-Weierstrass there is a convergent subsequence $x\_{n\_i}\to x\in[a,b]$. Now $g(x\_{n\_i})-f(x\_{n\_i})\to 0$, so $g(x)=f(x)$.