Prove that $\int\limits_{x}^{+\infty}\frac{ds}{1+s^2}\geq \frac{x}{1+x^2},~x\geq 0.$

Question

Prove that for all $x\geq 0$: $$\int\limits_{x}^{+\infty}\frac{ds}{1+s^2}\geq \frac{x}{1+x^2}.$$

Attempt. I have tried algebra, by using $f(x)=\int\limits_{x}^{+\infty}\frac{ds}{1+s^2}- \frac{x}{1+x^2}$, where $$f'(x)=-\frac{1}{1+x^2}-\bigg(\frac{x}{1+x^2}\bigg)'=\ldots=-\frac{2}{(1+x^2)^2}<0,$$ so $f$ is decreasing and: $$f(x)\leq f(0)=\frac{\pi}{2},$$ but i didn't get anywhere. Geometrically, I tried bu using the area under the graph of $\frac{1}{1+x^2}$ after $x$ and the rectangle over $[0,x]$, of height $\frac{1}{1+x^2}$, but i couldn't prove the desired inequality.

Thanks in advance.

Answer 1

You already showed that $f$ is (strictly) decreasing, therefore $$ f(x) > \lim_{t \to \infty} f(t) = 0 \, . $$

Alternatively, substitute $s = 1/u$: $$ \int_{x}^{+\infty}\frac{ds}{1+s^2} = \int_{0}^{1/x}\frac{du}{1+u^2} > \frac 1x \frac{1}{1+\frac{1}{x^2}} = \frac{x}{1+x^2} \, . $$

Answer 2

You can express the right hand side as a definite integral: $$\int\limits_{x}^{+\infty}\frac{ds}{1+s^2}\geq \frac{x}{1+x^2} \iff \int\limits_{x}^{+\infty}\frac{1}{1+s^2}ds\geq \int\limits_{x}^{+\infty}\frac{s^2-1}{(1+s^2)^2}ds \iff \\ \int\limits_{x}^{+\infty}\frac{1}{1+s^2}-\frac{s^2-1}{(1+s^2)^2}ds\geq 0 \iff \int\limits_{x}^{+\infty}\frac{2}{(1+s^2)^2}ds\geq 0. \qquad \checkmark$$