<s>[INST] Prove that $\int\limits_{x}^{+\infty}\frac{ds}{1+s^2}\geq \frac{x}{1+x^2},~x\geq 0.$ 
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.
 [/INST] You can express the right hand side as a definite integral:
$$\int\limits_{x}^{+\infty}\frac{ds}{1+s^2}\geq \frac{x}{1+x^2} \iff \int\limits_{x}^{+\infty}\frac{1}{1+s^2}ds\geq \int\limits_{x}^{+\infty}\frac{s^2-1}{(1+s^2)^2}ds \iff \\
\int\limits_{x}^{+\infty}\frac{1}{1+s^2}-\frac{s^2-1}{(1+s^2)^2}ds\geq 0 \iff \int\limits_{x}^{+\infty}\frac{2}{(1+s^2)^2}ds\geq 0. \qquad \checkmark$$
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