Prove that if $f'(x) > g'(x)$, $f(a) \ge g(a)$ then $f(x) > g(x)$ in $(a,b]$ There's a similar question here: If $f(a) = g(a)$ and $f&#39;(x) &lt; g&#39;(x)$ for all $x \in (a,b)$, then $f(b) &lt; g(b)$ however in the original answer the conditions are a bit different as well as the solution uses Mean value theorem which we didn't learn.


  $f(x)$ and $g(x)$ are continuous in $[a,b]$ and differentiable in $(a,b)$. $\quad f(a) \ge g(a)$ and $f'(x) &gt; g'(x)$ in $(a,b)$. Prove that $f(x) &gt; g(x)$ in $(a, b]$.


Let $E(x) = f(x) - g(x)$. 

If $E(a) = E(b)$ then by Rolle theorem there must be a $c$ such that $E'(c) = 0$. 

But $E'(x) = f'(x) - g'(x) &gt; 0$ for $a&lt;x&lt;b$ therefore $E'(x) \neq 0$. Therefore $E(a) \neq E(b)$. 

Because $E(x)$ is continuous in $[a,b]$ there must be an absolute minimum and maximum for $E(x)$ in $[a,b]$. According to max/min theorem possible max/min can be on the edges of an interval, where the derivative is $0$ or where derivative doesn't exist. Because $E(x)&gt;0$ and differentiable in $(a,b)$ the max/min has to be on the edges.

$E'(x) &gt; 0 \Rightarrow$ $E(x)$ is ascending for $a&lt;x&lt;b$ therefore $E(a)$ = min, $E(b)$ = max.

Because $f(a) \ge g(a)$ then $E(a) \ge 0$. Because $E(a)$ is the absolute min from there $E(b) &gt;E(x)&gt;E(a)$ for $a&lt;x&lt;b$. Of course there can't be any $x_1, x_2$ such that $E(x_1)=E(x_2)$ because then again by Rolle $E'(x)$
 will have a zero but this contradicts the given conditions. 

Therefore if $E(b) &gt;E(x)&gt;E(a)$ for $a&lt;x&lt;b$ then $f(x)&gt;g(x)$ for $a&lt;x&lt;b$. Q.E.D.

Please let me know if there were parts I could've skipped in the proof.
The fact that $E$ is strictly increasing, is usually proven by MVT. Actually you don't need MVT. 

Notice that for $x\in (a,b)$ we have
$$ 0 &lt; E'(x) = \lim_{h\to 0^+} \underbrace{\frac1h}_{&gt;0} (E(x+h) - E(x)). $$
That is, there exists some $h_x &gt; 0$ such that for every $h\in (0,h_x)$ it follows $E(x+h) &gt; E(x)$. Now let $y\in (x, b]$. If $E(y) &gt; E(x)$ there is nothing to show. If $E(y) \le E(x)$ you obtain a contradiction by Rolle's theorem. 

Then, by continuity of $E$ and for $t\in (0,1)$, $t\to 0$ we have
$$ E(y) &gt; E(x) &gt; E(a (1-t) + xt) \to E(a) \ge 0. $$
Everything is good except for one thing I believe, you said $f(x) &gt; g(x) \in (a, b]$. Why is it one open one closed?