We will show that the maximum value is $\frac{1}{16}$, obtained when $a = b = \frac{1}{2}$.

The inequality $E(a, b) \le \frac{1}{16}$ is equivalent to $16(a+b) \le (4a^2+3)(4b^2+3)$, which we rewrite as $(4ab-1)^2+4(a+b-1)^2+2(2a-1)^2+2(2b-1)^2 \ge 0$, obviously true.

We will show that the maximum value is $\frac{1}{16}$, obtained when $a = b = \frac{1}{2}$.

We combine the following inequalities: $a+b \le \frac{(1+a+b)^2}{4}$ and (from CBS) $(4a^2+3)(4b^2+3) = (4a^2+1+2)(1+4b^2+2) \ge (2a+2b+2)^2$. We get:
$$\frac{a+b}{(4a^2+3)(4b^2+3)} \le \frac{(a+b+1)^2}{4(4a^2+3)(4b^2+3)} \le \frac{(a+b+1)^2}{4(2a+2b+2)^2} \le \frac{1}{16}.$$