Maximum of $\frac{1+3a^2}{(a^2+1)^2}$ What is the maximum value of
$\displaystyle{{1 + 3a^{2} \over \left(a^{2} + 1\right)^{2}}}$, given that $a$ is a real number, and for what values of $a$ does it occur ?.

Let $x=a^2\ge0$.  We then have to look at
$$\frac{1+3x}{(x+1)^2}$$
upon differentiating and setting equal to $0$, we get
$$\frac{3(x+1)^2-2(x+1)(1+3x)}{(x+1)^4}\\\implies3(x+1)^2=2(x+1)(1+3x)\\\implies3(x+1)=2(1+3x)\\\implies x=\frac13$$
Thus, the relative maxima (or minima) occurs at
$$a=\pm\sqrt x=\pm\frac{\sqrt3}3$$
