# With this function $f:(a,b)\to \mathbb R$, is that true $b-a\geq\pi$?

I'm trying to solve this question and I don't even how to begin:

Let $f:(a,b)\to \mathbb R$ be differentiable such that $f'$ is continuous, $\lim_{x\to a^+}f(x)=+\infty$, $\lim_{x\to b^-}=-\infty$ and $f'(x)+f(x)^2\geq -1$, for $x\in (a,b)$. Prove that $b-a\geq\pi$

There is a hint we have to use the function $g(x)=x+\arctan f(x)$

I need some help or clue, everything would be useful.

thanks a lot!

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Hint: You can find that $g'(x)=1+\frac{f'(x)}{1+f^2(x)}\geq 0$ so it is a differentiable increasing function so $g(a^+)<g(b^-)$. Now use two limits to find $\lim_{x\to a^+} g(x)$ and $\lim_{x\to b^-}g(x)$ and then connect them.
what do you mean by $g(a^+)$? thank you for your answer! –  user42912 Feb 7 '13 at 3:57
@user42912: I mean $g(x),~~x=a+\epsilon,\epsilon>0$ –  Babak S. Feb 7 '13 at 4:02