Right derivative of continuous function nonnegative implies increasing function? Suppose $f$ is a continuous function on $(a,b)$ such that $$\lim_{h\to 0^+}\frac{f(x+h)-f(x)}{h}\geq 0$$ exists for all $x\in (a,b)$.
Prove that $f$ is an increasing function on $(a,b)$, i.e. $f(x_1)\geq f(x_0)$ for all $x_1\geq x_0$, $x_0,x_1\in (a,b)$.

What I tried is proof by contradiction: Suppose there exists $x_1> x_0$ but $f(x_1)<f(x_0)$.
By Intermediate Value Theorem, there exists $x_0<c_1<x_1$ such that $f(x_1)<f(c_1)<f(x_0)$.
By IVT again, there exists $x_0<c_2<c_1$ such that $f(c_1)<f(c_2)<f(x_0)$.
By repeated use of IVT, there exists $$x_0<\dots c_3<c_2<c_1$$ with $$f(c_1)<f(c_2)<f(c_3)<\dots<f(x_0)$$
Since $c_n$ is decreasing and bounded below, $L:=\lim c_n$ exists. Similarly, $\lim f(c_n)=f(L)$ exists.
Then $$\lim_{h\to 0^+}\frac{f(L+h)-f(L)}{h}=\lim_{n\to\infty}\frac{f(c_n)-f(L)}{c_n-L}\leq 0$$
since $f(c_n)-f(L)<0$ and $c_n-L>0$.
This is almost enough to be a contradiction had it been a strict $<0$. However at this point it seems not enough. 
Thanks for any help. Note: I wouldn't rule out the possibility that the question itself is wrong..
Update: I think I got it. First prove the statement for a function $g$ with $\frac{g(x+h)-g(x)}{h}\geq\epsilon>0$ on $(a,b)$. The above proof will do the job since now a contradiction can be achieved.
Then apply it to the function $g(x)=f(x)+\epsilon x$. Since $\epsilon>0$ is arbitrary, it can be proved that $f$ is increasing.

See snapshot of Hardy's book question 19: 
