Prove by Mean Value Theorem: $\frac15(b-a) < \arctan\left(1+\frac1a\right) - \arctan\left(1+\frac1a\right)$

Question

Prove $$\frac15(b-a) < \arctan\left(1+\frac1a\right) - \arctan\left(1+\frac1b\right) < b-a; \qquad\text{ for }0<a<b<1.$$ I tried using the fact that arctan function is monotonous increasing but I simply couldn't solve it. I know that it might be easier using Taylor series around x=0 but there has to be a way only using the MVT.

Answer 1

The mean-value theorem (MVT) states that under sufficient smoothness conditions for $f$ on the interval $(a,b)$, $0<a<b<1$, there exists a number $ \xi\in (a, b)$ such that

$$f(b)-f(a)=f'(\xi )(b-a)$$

Appliying the MVT to $f(x) =\arctan (1+\frac{1}{x})$ reveals that for some $ \xi\in (a,b)$

$$\arctan\left(1+\frac{1}{a}\right)-\arctan\left(1+\frac{1}{b}\right)=\frac{(b-a)}{\xi^2+(1+\xi)^2}$$

Then, since $0<\xi<1$, we have the inequalities

$$\frac15(b-a)<\frac{(b-a)}{\xi^2+(1+\xi)^2}<(b-a)$$