Maximum of $\frac{1+3a^2}{(a^2+1)^2}$

Question

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 ?.

Answer 1

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

$$a=\pm\sqrt x=\pm\frac{\sqrt3}3$$

Answer 2

Let $a^2=x$.

Hence, $\frac{1+3x}{(1+x)^2}\leq\frac{9}{8}$ it's $(3x-1)^2\geq0$.

The equality occurs for $x=\frac{1}{3}$, which says that the answer is $\frac{9}{8}$.

Answer 3

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 $\cfrac{3 \cdot 3}{4}- \cfrac{2 \cdot 9}{16} = \cfrac{9}{8}\,$.