Radicals Everywhere

Relevant wiki: Radical Equations - Intermediate

$\sqrt{a - 2} + \sqrt{b - a - 2} + \sqrt{c - b- 3} = \frac{1}{2}c - 2$

Multiplying by 2 we get:

$2\sqrt{a - 2} + 2\sqrt{b - a - 2} + 2\sqrt{c - b- 3} = c - 4$

$c - 4 - 2\sqrt{a - 2} - 2\sqrt{b - a - 2} - 2\sqrt{c - b- 3} = 0$

$(a - 2 - 2\sqrt{a - 2} + 1) + (b - a - 2 - 2\sqrt{b - a - 2} + 1) + (c - b - 3 - 2\sqrt{c - b- 3} + 1) = 0$

$(\sqrt{a - 2} - 1) ^ {2} + (\sqrt{b - a - 2} - 1) ^ 2 + (\sqrt{c - b- 3} - 1) ^ 2 = 0$

a - 2 = 1 ==> a = 3

b - a - 2 = 1 ==> b = 6

c - b - 3 = 1 ==> c = 10

$\sqrt{2b + 3b + 4c} = \sqrt{6 + 18 + 40} = 8$

Answer = 8

Relevant wiki: Applying the Arithmetic Mean Geometric Mean Inequality

It is clear that $a-2$ , $b-a-2$ and $c-b-3$ are non-negative,

Then applying AM-GM on $(a-2)$ and 1 we have $\frac{a-1}{2}\geq\sqrt{a-2}$

Now apply AM-GM on $(b-a-2)$ and 1 to get $\frac{b-a-1}{2}\geq\sqrt{b-a-2}$

And finally AM-GM on $(c-b-3)$ and 1 to get $\frac{c-b-2}{2}\geq\sqrt{c-b-3}$

Adding all three we have $\sqrt{a-2}+\sqrt{b-a-2}+\sqrt{c-b-3}\leq\frac{1}{2}c-2$

The equation asks us for strict equality which for AM-GM only happens when all the terms are equal so,

$a-2=1 \implies a=3$

$b-a-2=1 \implies b=6$

$c-b-3=1 \implies c=10$

Thus $\sqrt{2a+3b+4c}=8$