Classic Algebra Question I

By taking square for both sides,

$\begin{aligned} a+\sqrt{c}+2\sqrt{b+\sqrt{c}} & = 6+2\sqrt{6}+2\sqrt{2}+2\sqrt{3} \\ & = 6+2\sqrt{6}+2\sqrt{(2+3)+2\sqrt{(2)(3)}} \\ & = 6+\sqrt{24}+2\sqrt{5+\sqrt{24}} \\ \end{aligned}$

Clearly see that $a=6, b=5, c=24.$

Thus, we will obtain $a+b+c=\fbox{35}.$

$\begin{aligned} \sqrt{a+\sqrt c+2\sqrt{b+\sqrt c}} & = 1 + \sqrt 2 + \sqrt 3 & \small \color{#3D99F6} \text{Squaring both sides} \\ a+\sqrt c+2\sqrt{b+\sqrt c} & = 1 + 2 + 3 + 2\left(\sqrt 2 + \sqrt 6+ \sqrt 3\right) \\ & = 6 + 2\sqrt 6 + 2\sqrt{\left(\sqrt 2 + \sqrt 3\right)^2} \\ & = 6 + 2\sqrt 6 + 2\sqrt{5 + 2\sqrt 6} \\ & = 6 + \sqrt{24} + 2\sqrt{5 + \sqrt{24}} \end{aligned}$

Therefore, $a+b+c = 6+5+24 = \boxed{35}$ .