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 ?.

Writing it as $\cfrac{1+3a^2}{(a^2+1)^2}= \cfrac{3(a^2+1)-2}{(a^2+1)^2}= \cfrac{3}{a^2+1}-\cfrac{2}{(a^2+1)^2}\,$ gives a quadratic in  $x=\cfrac{1}{a^2+1}\,$, with a maximum at $x = \cfrac{3}{4}\,$ ($\iff a^2 = 1/3\,$) of value $\cfrac{3 \cdot 3}{4}- \cfrac{2 \cdot 9}{16} = \cfrac{9}{8}\,$.
