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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.

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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 $x \geq x_0$.

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  • $\begingroup$ just a quick question, what is the difference between $h \equiv 0$ and $h = 0?$ $\endgroup$ – Donlans Donlans Jul 30 '19 at 23:32
  • $\begingroup$ When I say $h \equiv 0$ in $[a,x_0]$ I mean that $h(x)=0$ for all $x \in [a,x_0]$. $\endgroup$ – Kavi Rama Murthy Jul 30 '19 at 23:33
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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 $\tilde x < x_0$.

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