<s>[INST] Show that $\frac{\arctan(x^2)}{\ln(1+x^2)}$ is bounded by $\frac{1+\sqrt{2}}{2}$ Given $F(x) = \frac{\arctan(x^2)}{\ln(1+x^2)},$ which is defined for $x>0.$ I would like to show that it is bounded above by $\frac{1+\sqrt{2}}{2}$.
I tried to differentiate the function in order to find maximal points however I've reached an equation which I don't know how to solve :
$$F'(x) = \frac{\frac{2x}{1+x^4}\ln(1+x^2)-\arctan(x^2)\frac{2x}{1+x^2}}{\ln^2(1+x^2)}\overset{{\rm ?}}{=}0$$
Any help would be appreciated.
 [/INST] We can simplify the problem by treating the function in terms of $x^2$. That is, the maximum value of $\arctan(x^2)/\log(1+x^2)$ is the same as that of $\arctan x/\log(1+x)$ which has derivative $$\frac1{\log^2(1+x)}\left(\frac{\log(1+x)}{1+x^2}-\frac{\arctan x}{1+x}\right).$$ Setting this to zero yields an expression for the stationary point $x_s$ $$\frac{\log(1+x_s)}{1+x_s^2}=\frac{\arctan x_s}{1+x_s}\implies\frac{\arctan x_s}{\log(1+x_s)}=\frac{1+x_s}{1+x_s^2}.$$ To find the maximum value of $(1+x)/(1+x^2)$, we set its derivative $(1-2x-x^2)/(1+x^2)^2$ to zero which yields $x=-1\pm\sqrt2$. We know that $x_s>0$ (since the original function is a function of $x^2$) so it follows that $$\frac{\arctan x^2}{\log(1+x^2)}\le\frac{1+(-1+\sqrt2)}{1+(-1+\sqrt2)^2}=\frac{1+\sqrt2}2$$ as desired.
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