# Let $f(a) = g(a)$, $f'(x) \geq g'(x)$ for all x and that $f'(x_0) > g'(x_0)$ for some $x_0 > a$ show that $f(x) > g(x)$ for all $x \geq x_0$

By the mean value theorem, we have $$f(x) \geq g(x)$$ for all $$x > a$$, after that, I'm stuck, I've tried several things, like adding the inequalities, working with the definition of derivative, etc.

Could someone give me hint?. I don't want the solution, just a hint.

Let $$h(x)=f(x)-g(x)$$. If $$h(x_0)=0$$ then $$h \equiv 0$$ in $$[a,x_0]$$ because $$h' \geq 0$$ (so $$h$$ is non -decreasing) and $$h(a)=h(x_0)=0$$. But this contradicts the hypothesis that $$h'(x_0)>0$$. It follows that $$h(x_0) >0$$. [Note that $$h(x_0)=h(a)+h'(\xi)$$ for some $$\xi \in (a,x_0)$$ so $$h (x_0) \geq 0$$]. Now $$h(x) -h(x_0)=(x-x_0)h'(t)$$ for some $$t$$ which gives $$h(x) \geq h(x_0) >0$$ for all $$x >x_0$$. Hence $$f(x) > g(x)$$ for all $$x \geq x_0$$.
• just a quick question, what is the difference between $h \equiv 0$ and $h = 0?$ – Donlans Donlans Jul 30 '19 at 23:32
• When I say $h \equiv 0$ in $[a,x_0]$ I mean that $h(x)=0$ for all $x \in [a,x_0]$. – Kavi Rama Murthy Jul 30 '19 at 23:33
A rough hint would be if $$f'(x) \geq g'(x)$$ for all $$x \geq a$$, then you would have $$f(x) \geq g(x)$$ for all $$x > a$$ by the mean value theorem. However, since you have $$f'(x_0) > g'(x_0)$$ for some $$x_0 > a$$, I think you should apply the mean value theorem over 2 intervals, $$[a, \tilde x]$$ and $$[\tilde x, x]$$, where $$\tilde x < x_0$$.