For $f(x)$ on $(a,b)$ s.t. $f'(x)+f^2(x) \ge -1$ and $\lim\limits_{x \to a} f(x) = - \lim\limits_{x \to b} f(x) = \infty$, prove that $b-a \ge \pi$

Question

$f(x)$ is continuously differentiable here. Using separation of variables, I think I might have shown that the equality form of the statement is true, but I'm a bit wary of trying separation of variables for a differential inequality. There was no solution provided for the problem.

Answer 1

Note that from $f'(x)+f^2(x) \ge -1$, we have $$ f'(x)\ge -(f^2(x)+1) $$ or $$ \frac{df(x)}{f^2(x)+1}\ge -dx. $$ Integrating both sides from $a+\varepsilon$ to $b-\varepsilon'$ ($b-a>\varepsilon'>\varepsilon>0$) yields $$ \int_{a+\varepsilon}^{b-\varepsilon'}\frac{df(x)}{f^2(x)+1}\ge -\int_{a+\varepsilon}^{b-\varepsilon'}dx. $$ So $$ \arctan f(b-\varepsilon')-\arctan f(a+\varepsilon)\ge-(b-\varepsilon'-a+\varepsilon)$$ or $$ b-a-\varepsilon'+\varepsilon\ge\arctan f(a+\varepsilon)-\arctan f(b-\varepsilon').$$ Letting $\varepsilon,\varepsilon'\to 0^+$ yields the desired inequality.