Prove that (b-a)/5 < arctan (1+ dfrac1a)- arctan (1+

Prove that $\dfrac{b-a}{5} \lt \arctan\!\left(1+\dfrac1a\right)-\arctan\!\left(1+\dfrac1b\right) \lt b-a$ for $0\lt a\lt b\lt 1$.

Step 1: The function to differentiate

\[ f(x) = \arctan\!\left(1+\frac1x\right), \qquad x>0, \]

so the quantity to be bounded is $f(a)-f(b)$.

Step 2: The derivative simplifies remarkably

With $u = 1+\dfrac1x$ and $u' = -\dfrac{1}{x^{2}}$: \[1+u^{2} = 1+\left(1+\frac1x\right)^{2} = 2+\frac2x+\frac{1}{x^{2}} = \frac{2x^{2}+2x+1}{x^{2}},\]

\[ f'(x) = \frac{x^{2}}{2x^{2}+2x+1}\cdot\left(-\frac{1}{x^{2}}\right) = -\frac{1}{2x^{2}+2x+1}. \]

The factor $x^{2}$ cancels completely, leaving a polynomial reciprocal.

Step 3: Apply Lagrange's theorem

On $[a,b]$ there is a $\xi\in(a,b)\subset(0,1)$ with

\[ f(b)-f(a) = -\frac{b-a}{2\xi^{2}+2\xi+1} \;\Longrightarrow\; f(a)-f(b) = \frac{b-a}{2\xi^{2}+2\xi+1}. \tag{1} \]

Step 4: Bound the denominator

The quadratic $q(\xi) = 2\xi^{2}+2\xi+1$ has $q'(\xi) = 4\xi+2>0$ on $(0,1)$, so it increases from $q(0) = 1$ to $q(1) = 5$: $1 \lt 2\xi^{2}+2\xi+1 \lt 5 \ \text{ for } \xi\in(0,1)$.

Step 5: Invert and conclude

Taking reciprocals reverses the inequalities, and multiplying by $b-a>0$ gives, with (1), $\frac{b-a}{5} \lt \arctan\!\left(1+\frac1a\right)-\arctan\!\left(1+\frac1b\right) \lt b-a$.

Step 6: Verification with $a = \dfrac12,\ b = \dfrac34$

\[b-a = 0.25:\quad \text{left } = 0.05, \qquad \text{right } = 0.25\]

\[\arctan3-\arctan\frac73 = 1.249046-1.165905 = 0.083141\]

\[0.05 \lt 0.083141 \lt 0.25 \;\;\checkmark\]