<s>[INST] A continuous function $f:\left[-\frac{\pi}{4},\frac{\pi}{4}\right]\to[-1,1]$ and differentiable on $\left(-\frac{\pi}{4},\frac{\pi}{4}\right)$. 

$\blacksquare~$ Problem: Suppose a continuous function $f:\left[-\frac{\pi}{4},\frac{\pi}{4}\right]\to[-1,1]$ and differentiable on $\left(-\frac{\pi}{4},\frac{\pi}{4}\right)$. Then, there exists a point $x_0\in \left(-\frac{\pi}{4},\frac{\pi}{4}\right)$ such that
$$|f'(x_0)|\leqslant 1+f(x_0)^2$$

$\blacksquare~$ My Solution: Let's take $g(x) = \tan^{-1} f(x) $. Then $g : \left[ - \frac{\pi}{4}, \frac{\pi}{4}\right] \to \left[- \frac{\pi}{4}, \frac{\pi}{4} \right] $.
Now, as $f$ is cont in $\left[- \frac{\pi}{4}, \frac{\pi}{4} \right]$ and differentiable in $\left(- \frac{\pi}{4}, \frac{\pi}{4} \right)$, $g$ is also the same.
By LMVT, we have that
$$\frac{g\left(\frac{\pi}{4}\right) - g\left(-\frac{\pi}{4} \right) }{\frac{\pi}{2}} = g'(x_0) \quad \text{for some } x_0 \in \left(- \frac{\pi}{4}, \frac{\pi}{4} \right)$$$$\implies  \frac{ \frac{\pi}{4} - \left(- \frac{\pi}{4}\right) }{ \frac{\pi}{2} } \geqslant \frac{g\left(\frac{\pi}{4}\right) - g\left(-\frac{\pi}{4} \right) }{\frac{\pi}{2}} = \left(\tan^{-1}f(x_0) \right)' = \frac{f'(x_0)}{1 + f(x_0)^2} $$$$ \implies 1 + f(x_0)^2 \geqslant f'(x_0) $$Again, after the LMVT part, we have that
$$ \implies  \frac{ - \frac{\pi}{4} - \left( \frac{\pi}{4}\right) }{ \frac{\pi}{2} } \leqslant \frac{g\left(\frac{\pi}{4}\right) - g\left(-\frac{\pi}{4} \right) }{\frac{\pi}{2}} = \left(\tan^{-1}f(x_0) \right)' = \frac{f'(x_0)}{1 + f(x_0)^2} $$$$ \implies - \left( 1 + f(x_0)^2 \right) \leqslant f'(x_0) $$Hence, combining these two we have that$$ \lvert f'(x_0) \rvert \leqslant 1 + f(x_0)^2 \quad \text{for some } x_0 \in \left( - \frac{\pi}{4}, \frac{\pi}{4} \right) $$


Is this fine? Is there any glitch? Another way of a solution will be great!
Regards, Ralph
 [/INST] Your solution is fine, you may just simpify a bit just one step:
You can write from the begging
$\left\vert \frac{g\left(\frac{\pi}{4}\right) - g\left(-\frac{\pi}{4} \right) }{\frac{\pi}{2}}\right\vert = \vert g'(x_0)\vert  \quad \text{for some } x_0 \in \left(- \frac{\pi}{4}, \frac{\pi}{4} \right)$ so you can get immediately
$\vert g'(x_0)\vert \le 1$.
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