<s>[INST] How do I know that $\ln(x^2+1)-x \arctan(x)$ is always negative or zero? I did google the function and I can clearly see that it is always negative or zero, but I have no idea how I would have found this on my own. Both the logarithm and the $x\cdot \arctan(x)$ are positive.
What I did do is derive the function but that did not seem to be more obvious.
$\rho' = \frac{x}{1+x^{2}}-\arctan(x)$ 
This also does not look obvious to me.
$\rho'' = -\frac{2\cdot x^{2}}{(1+x^{2})^{2}}$
This is most definitely negative, thus concave (or concave downwards). Is that enough to deduce anything ?Am I missing something?
 [/INST] let $$y = \ln(1 + x^2 ) - x \tan^{-1} x .$$  the function is even and $y = 0$  at $x = 0 \text{ and  } \lim_{x \to \infty} y = -\infty. $  taking the derivative, you find $$\frac{dy}{dx} = \frac{2x}{1+x^2} - \frac{x}{1+x^2} - \tan^{-1} x =\frac{x}{1+x^2} - \tan^{-1} x$$ we will have established that $y < 0 \text{ for all } x \neq 0$ if we can show that $$f(x) = \frac{x}{1+x^2} - \tan^{-1} x <0, x \neq 0.$$ observe $x = 0$ is the only zero of $f.$ suppose $$x > 0, f'(x) = \frac{1}{1+x^2} - \frac{2x^2}{1+x^2} -\frac 1{1+x^2} = -\frac{2x^2}{1+x^2} < 0$$  together with $f(0) = 0$ proves the claim and we are done.
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