TITLE: If $f(a) = g(a)$ and $f'(x) < g'(x)$ for all $x \in (a,b)$, then $f(b) < g(b)$
QUESTION [9 upvotes]: Assume that $f$ and $g$ are continuous on $[a, b]$ and differentiable on $(a, b)$.
Prove that if $f(a) = g(a)$ and $f'(x) < g'(x)$ for all $x \in (a,b)$, then $f(b) < g(b)$.

I understand that if $f$ and $g$ start at the same point, and $g$ increases at a faster rate than $f$, it will have a bigger end value. I'm just not sure how to prove it.
Consider $h(x)=  g(x) - f (x)$ (this is the space between the two functions)
$$\begin{align*}
h(a) &= g(a) - f(a)\\
h(a) &= 0 &(\text{they are in the same spot in the beginning})
\end{align*}$$
$$\begin{align*}
h'(x) &= \frac{g'(x) - f'(x)}{g(x) - f(x)}\\
h'(x) &> 0 &(\text{since } g'(x)> f'(x))
\end{align*}$$
so $g(x) - f(x) > 0 \implies g(x) > f(x)$?
Is this correct?

REPLY [5 votes]: Letting $h = g - f$ is a good first step. Now apply the Mean Value Theorem to $h$ to note that for some $x \in [a, b]$, $$h'(x) = \frac{h(b)-h(a)}{b-a},$$ or equivalently $$g'(x)-f'(x)=\frac{g(b) - f(b) - g(a) + f(a)}{b-a}.$$
Because we've assumed that $g'(x) > f'(x)$, we know the $g'(x) - f'(x)$ is positive, and so therefore so is the right hand side of this equation. So we now get
$$0 < \frac{g(b)-f(b)-g(a)+f(a)}{b-a} \implies0<g(b)-f(b)-g(a)+f(a).$$
Since we know that $g(a) = f(a)$, we can now say $$0<g(b)-f(b) - f(a)+f(a) \implies 0 <g(b)-f(b)\implies f(b) < g(b). \square$$