Suppose that $\,f:[a,b]\to \mathbb{R}$, where $\,b-a\ge 4,\,$ is differentiable in $(a,b)$ and continuous in $[a,b]$. Prove that there is $x_0\in (a,b)$, such that
$$f'(x_0)<1+\big(\,f(x_0)\big)^2\!.$$
But, I could not make the slightest approach towards the solution of this problem. Please help. Thank you.
First proof. (Assuming that $\,f$ is continuously differentiable). Assume that for every $x\in[a,b]$: $$ f'(x)\ge 1+f^2(x), $$ then $$ \frac{d}{dx}\tan^{-1}\big(f(x)\big)=\frac{f'(x)}{1+f^2(x)}\ge 1, $$ and thus integrating in $[a,b]$ $$ \tan^{-1}\big(f(b)\big)-\tan^{-1}\big(f(a)\big)\ge b-a\ge 4. $$ But $\tan^{-1}(x)\in(-\pi/2,\pi/2)$, and hence $$ \tan^{-1}\big(f(b)\big)-\tan^{-1}\big(f(a)\big)<\pi. $$
Alternative proof. (Assuming that $f$ is ONLY differentiable). As $\tan^{-1}: \mathbb R\to (-\pi/2,\pi/2)$, then, using the mean value theorem for $g(x)=\tan^{-1}\big(f(x)\big)$, with $g'(x)=f'(x)/\big(1+f^2(x)\big)$, we obtain that there exists a $\xi\in (a,b)$, such that $$ \pi>\tan^{-1}\big(f(b)\big)-\tan^{-1}\big(\,f(a)\big)= (b-a)\frac{f'(\xi)}{1+f^2(\xi)}, $$ and hence $$ 1+f^2(\xi)>\frac{\pi}{b-a}\big(1+f^2(\xi)\big)= f'(\xi). $$