Prove if $f(a)=g(a)$ and $f′ >g$′ on $(a,b)$, then $f>g$ on $(a, b)$. Where f, g are differentiable and $a<b$.

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Suppose f and g be differentiable on an open interval. Let $a, b \in$ the interval and let $a<b$. Prove if $f(a)=g(a)$ and $f′ >g$′ on $(a,b)$, then $f>g$ on $(a b)$.

Up to this point I have Rolle's theorem, Cauchy mean value theorem, and the mean value theorem. I know that since a, b are in the open interval, f and g are continuous on $[a,b]$ and differentiable on $(a,b)$.

Visually/ conceptually I understand this, but am not sure how to prove it. I think the proof involves Rolle's theorem, but I can't manipulate it to work for all $x\in (a,b)$.

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Hint: Use the Mean Value Theorem on $h = f - g$.