Nice usage

By RMS - AM inequality,

$\sqrt{\dfrac{\sqrt{4a+1}^2 + \sqrt{4b+1}^2}{2}} \geq \dfrac{\sqrt{4a+1} + \sqrt{4b+1}}{2}$

$\sqrt{\dfrac{4a + 1 + 4b + 1}{2}} \geq \dfrac{\sqrt{4a+1} + \sqrt{4b+1}}{2}$

$2 \sqrt{\dfrac{6}{2}} \geq \sqrt{4a+1} + \sqrt{4b+1}$

$2\sqrt{3} \geq \sqrt{4a+1} + \sqrt{4b+1}$

Equality occurs when $a = b = \frac{1}{2}$ .

It is symmetric with respect to the two variables in both expressions. So, a=b =1/2.
So, exp. = $\sqrt{4*1/2 +1} + \sqrt{4*1/2 +1} = 2*\sqrt3= \boxed{ 3.4641 }$
OR $\\~~ \\ exp.= \sqrt{4*a +1} + \sqrt{4 -4*a +1} =\sqrt{4*a +1} + \sqrt{5 -4*a} \\ Differantiating~~ w.r.t~~.a,~ and~ equating~ to~ 0.\\We~get ~\dfrac{2}{\sqrt{4a+1} } = \dfrac{2}{\sqrt{5 -4a} } ~\\\implies~ 4a+1=5 -4a.....gives~~a=1/2 \therefore~b=1/2,~~~\\and~ exp.= \boxed { 3.4641 }$

No need of complex equations.This needs only maxima and minima here's the solution:

using the Cauchy-Schwarz inequality: let Y1 = Y2 = 1, X1 = $\sqrt{4a+1}$ and X2 = $\sqrt{4b+1}$ . Substituting this gives us: ( $\sqrt{4a+1}$ + $\sqrt{4b+1}$ )^2 has a maximum value of 12 by applying a+b=1. square - rooting gives a maximum value of $\sqrt{12}$ = 2 $\sqrt{3}$ ~ 3.464. Equality holds if, and only if, (X1)(Y2) = (X2)(Y1) or in other words a=b=0.5