Radical Bound

Using Cauchy-Schwarz inequality , we have:

$\begin{aligned} (\sqrt{4a+1} + \sqrt{4a+1})^2 & \le (1+1)(4a+1+4b+1) = 8(a+b)+4 = 8(1) + 4 = \boxed{12} \end{aligned}$

Note that equality occurs when $a=b=\frac 12$ .