# Proving that $b-a\ge \pi$

Let $f : C^1(a; b)$, such that $\lim\limits_{x\to a^+} f(x) = +\infty$, $\lim\limits_{x\to b^-}f(x) = -\infty$ and $f'(x)+f^2(x) \ge -1$ for $x \in (a; b)$.

Prove that $b-a\ge \pi$ and provide an example where $b-a= \pi$

For the second question the obvious example could be $f(x) = \cot(x)$ with $a=0$, and $b=\pi.$

Any hint for the first part? This queston is similar to this: How prove this inequality $b-a\ge \pi$

Since $f^{2}+1$ is never zero, the inequality transforms into
$$\frac{f'(x)}{1+f^{2}(x)}\geq -1$$ We integrate this inequality from $a$ to $b$, and the LHS becomes \begin{align} \int_{a}^{b}\frac{f'(x)}{1+f^{2}(x)}\text{d} &= \int_{a}^{b}(\arctan{f(x)})'\text{d} x\\ &=-\frac{\pi}{2} -( \frac{\pi}{2})\\ &=-\pi \end{align} The RHS of the inequality is just $-(b-a)$, therefore $$-\pi\geq -(b-a)\Rightarrow b-a\geq \pi$$