Prove that ln (1+x) > (arctan x)/(1+x) for x > 0

Prove that $\ln\bigl(1+x\bigr)\gt \dfrac{\arctan x}{1+x}$ for $x\gt 0$.

Step 1: The auxiliary function

\[ f(x) = \ln\bigl(1+x\bigr)-\frac{\arctan x}{1+x}, \qquad x\ge0, \qquad f(0) = 0-0 = 0. \]

Step 2: Differentiate

For the quotient, \[\frac{d}{dx}\left[\frac{\arctan x}{1+x}\right] = \frac{\dfrac{1+x}{1+x^{2}}-\arctan x}{\bigl(1+x\bigr)^{2}},\] so, putting everything over $\bigl(1+x\bigr)^{2}$, \[f'(x) = \frac{\bigl(1+x\bigr)-\dfrac{1+x}{1+x^{2}}+\arctan x}{\bigl(1+x\bigr)^{2}} = \frac{\bigl(1+x\bigr)\left(1-\dfrac{1}{1+x^{2}}\right)+\arctan x}{\bigl(1+x\bigr)^{2}}.\]

Step 3: Simplify and read off the sign

Since $1-\dfrac{1}{1+x^{2}} = \dfrac{x^{2}}{1+x^{2}}$, \[f'(x) = \frac{\dfrac{x^{2}\bigl(1+x\bigr)}{1+x^{2}}+\arctan x}{\bigl(1+x\bigr)^{2}}. \tag{1}\]

For $x\gt 0$ both terms in the numerator are strictly positive and the denominator is positive, so $f'(x)\gt 0$.

Step 4: Conclusion

$f$ is strictly increasing on $\bigl[0,\infty\bigr)$ and $f(0) = 0$, hence $f(x)\gt 0$ for every $x\gt 0$.

Step 5: Verification

\[x = 0.1:\ 0.0953102 \gt 0.0906079; \qquad x = 1:\ 0.6931472 \gt 0.3926991 \;\;\checkmark\]

\[x = 5:\ 1.7917595 \gt 0.2289001 \;\;\checkmark\ \text{(the right side eventually decays)}\]

\[\text{Near } 0:\ \ln(1+x) = x-\frac{x^{2}}{2}+\cdots,\quad \frac{\arctan x}{1+x} = x-x^{2}+\cdots,\ \text{difference } \frac{x^{2}}{2}\gt 0 \;\;\checkmark\]