Prove $\exists \xi \in (a,b):f'(\xi)<1+f^2(\xi)$

Question

Problem

$f(x)$ is defined over $[a,b]$ and differentiable over $(a,b)$, where $b-a\geq 4.$Prove that there exists $\xi \in (a,b)$ such that $f'(\xi)<1+f^2(\xi)$.

My Proof

Since $b-a \geq 4$，we can obtain $$\exists x_1,x_2 \in (a,b):x_2-x_1>\pi.$$ Denote$$F(x):=\arctan f(x).$$ Obviously, $F(x)$ is continuous over $[x_1,x_2]$ and differentiable over $(x_1,x_2)$. Thus, By Lagrange's Mean Value Theorem，$$\exists \xi \in (x_1,x_2) \subset (a,b):F(x_2)-F(x_1)=F'(\xi)(x_2-x_1).$$ Further, $$\frac{f'(\xi)}{1+f^2(\xi)}=F'(\xi)=\frac{F(x_2)-F(x_1)}{x_2-x_1}\leq \frac{|F(x_2)|+|F(x_1)|}{x_2-x_1}<\frac{\frac{\pi}{2}+\frac{\pi}{2}}{\pi}=1,$$ Which implies $$f'(\xi)<1+f^2(\xi).$$

AM I RIGHT? HOPE TO SEE OTHER PROOFS. THX.

Answer 1

You are right. But you do not have to invoke $x_2,x_1,....$. Let $F$ as above.

By the mean value theorem there $\exists \xi \in (a,b)$ such that

$\frac{f'(\xi)}{1+f^2(\xi)}=\frac{F(b)-F(a)}{b-a}$

Then $\frac{f'(\xi)}{1+f^2(\xi)} \le \frac{|f'(\xi)|}{1+f^2(\xi)}= \frac{|F(b)-F(a)|}{b-a} \le \pi \frac{1}{b-a}<1$, since $b-a \ge 4$.

Answer 2

I would just word it in a more concise form, like...

If we assume that for any $\xi\in(a,b)$ the inequality $\frac{f'(\xi)}{1+f^2(\xi)}\geq 1$ holds, by termwise integration we have that $\arctan f(b)-\arctan f(a) \geq b-a \geq 4$. This is absurd since the LHS is at most $\pi$.