1. **Problem statement:** Show that for all $x > 0$, the inequality $$\frac{x}{1+x^2} < \tan^{-1} x < x$$ holds. 2. **Recall the functions and their properties:** - $\tan^{-1} x$ is the inverse tangent function, increasing and differentiable for $x > 0$. - We want to compare $\tan^{-1} x$ with $\frac{x}{1+x^2}$ and $x$. 3. **Step 1: Show $\tan^{-1} x < x$ for $x > 0$** - Define $h(x) = x - \tan^{-1} x$. - Compute derivative: $$h'(x) = 1 - \frac{1}{1+x^2} = \frac{x^2}{1+x^2} > 0 \text{ for } x > 0.$$ - Since $h'(x) > 0$, $h(x)$ is strictly increasing on $(0, \infty)$. - At $x=0$, $h(0) = 0 - 0 = 0$. - For $x > 0$, $h(x) > h(0) = 0$, so $x - \tan^{-1} x > 0$ or $$\tan^{-1} x < x.$$ 4. **Step 2: Show $\frac{x}{1+x^2} < \tan^{-1} x$ for $x > 0$** - Define $k(x) = \tan^{-1} x - \frac{x}{1+x^2}$. - Compute derivative: $$k'(x) = \frac{1}{1+x^2} - \frac{(1+x^2) - 2x^2}{(1+x^2)^2} = \frac{1}{1+x^2} - \frac{1 - x^2}{(1+x^2)^2}.$$ - Simplify numerator: $$k'(x) = \frac{(1+x^2)(1+x^2) - (1 - x^2)}{(1+x^2)^2} = \frac{(1+x^2)^2 - (1 - x^2)}{(1+x^2)^2}.$$ - Expand numerator: $$(1+x^2)^2 = 1 + 2x^2 + x^4,$$ so numerator is $$1 + 2x^2 + x^4 - 1 + x^2 = 3x^2 + x^4 = x^2(3 + x^2) > 0 \text{ for } x > 0.$$ - Therefore, $k'(x) > 0$ for $x > 0$, so $k(x)$ is strictly increasing. - At $x=0$, $k(0) = 0 - 0 = 0$. - For $x > 0$, $k(x) > 0$, so $$\tan^{-1} x > \frac{x}{1+x^2}.$$ 5. **Conclusion:** Combining both inequalities, for all $x > 0$, $$\frac{x}{1+x^2} < \tan^{-1} x < x.$$