Inequality to be proved

$\begin{aligned} \frac a{a^2+3} + \frac b{b^2+3} & = \frac {ab}{a^2b+3b} + \frac {ab}{ab^2+3a} & \small \blue{\text{Given that }ab = 1} \\ & = \frac 1{a+3b} + \frac 1{b+3a} \\ & = \frac 1{\blue{a+b+b+b}} + \frac 1{\blue{b+a+a+a}} & \small \blue{\text{By AM-GM inequality}} \\ & \le \frac 1{\blue{4\sqrt b}} + \frac 1{\blue{4 \sqrt a}} = \frac 12 = \boxed{0.5} & \small \blue{\text{Equality occurs when }a=b=1} \end{aligned}$

Reference: AM-GM inequality

Substituting $b=\dfrac{1}{a}$ in the given expression we see that the expression simplifies to $\dfrac{4a^3+4a}{3a^4+10a^2+3}$ . Applying the condition for optima we get that the expression attains a minimum at $a=b=-1$ , the minimum value of the expression being $-\dfrac{1}{2}$ , and a maximum at $a=b=1$ , the maximum being $\dfrac{1}{2}$

I tried it with MTH theorem and rearrangement but couldn't reach anywhere