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?
 A: For any $x\neq 0$ we have:
$$ \frac{d^2}{dx^2}\left(\log(1+x^2)-x\arctan x\right) = -\frac{2x^2}{(1+x^2)^2}<0\tag{1}$$
that proves concavity, hence the graphics of $\log(1+x^2)-x\arctan x$ always lie below the tangent in $x=0$, that is just the $x$-axis, since $\log(1+x^2)-x\arctan x$ is an even $C^\infty$ function whose value in zero is zero.
A: Let $f(x)=\ln(x^2+1)-x\arctan x$.  Since $f(0)=0$ and $f(-x)=f(x)$, it suffices to show that $f'(x)\lt0$ for $x\gt0$.  So noting
$$f'(x)={2x\over x^2+1}-\arctan x-{x\over x^2+1}={x\over x^2+1}-\arctan x$$
and
$$\arctan x=\int_0^x{dt\over t^2+1}\gt\int_0^x{dt\over x^2+1}={x\over x^2+1}$$
does the trick.
Remark:  This answer doesn't address the OP's question about the second derivative.  That portion was edited in while I was thinking and typing.  The version I was working from ended at "obvious."
A: 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.
