$f, g$ continuous on closed finite interval, $f(x) > g(x)$ for all $x$ from the interval, implies $f(x) > g(x) + \alpha$ for some $\alpha > 0$?

Question

The question:

Let $f(x)$ and $g(x)$ be continuous on the closed, finite interval $[a, b]$ with $f(x) > g(x)$ for all $x \in [a, b]$. Prove that there exists an $\alpha > 0$ such that $f(x) > \alpha + g(x)$ for all $x \in [a, b]$.

I tried using $\epsilon-\delta$ continuity as well as sequential continuity but didn't get anywhere. Any suggestions would be very helpful!

(Note: these are real-valued functions only.)

Answer 1

Hint: Does $\alpha = \dfrac{\min\{f(x)-g(x)\}}{2}$ satisfy $f(x) > g(x) + \alpha$ for $x \in [a,b]$?

Answer 2

Hint. It is important that this is for functions on a closed and bounded interval. For instance, $2x > x$ on $(0,1)$, but no constant $\alpha>0$ can exist so that $2x > \alpha+ x$ on $(0,1)$.