Hello trig, my old friend

Using $Cauchy-Schwarz \space Inequality$ we can easily show that $\textcolor{#3D99F6}{r.m.s. value}$ of $n$ positve numbers is greater than or equal to their $\space \textcolor{#3D99F6}{arithmetic \space mean}$

i.e. $\quad \LARGE \sqrt{\frac{a_{1}^{2} + a_2^2 + a_3^2 ... a_n^2}{n}} \geq \frac{a_1 + a_2 + a_3 ... a_n}{n}$

Now taking $n = 3$ and $a_1 = \sqrt{\sin\alpha\sin\beta\cos\gamma} , \space a_2 = \sqrt{\sin\beta\sin\gamma\cos\alpha} , \space a_3 = \sqrt{\sin\gamma\sin\alpha\cos\beta}$ We have

$\LARGE \frac{S}{3} \leq \sqrt{\frac{\sin\alpha\sin\beta\cos\gamma + \sin\beta\sin\gamma\cos\alpha + \sin\gamma\sin\alpha\cos\beta}{3}} \quad \small \cdots \text{Eq 1}$

Now to find maximum value of $S$ we have to find maximum value of RHS . Let $A$ denote $\sin\alpha\sin\beta\cos\gamma + \sin\beta\sin\gamma\cos\alpha + \sin\gamma\sin\alpha\cos\beta$

Using properties of $\sin \text{and} \cos$ and taking $\alpha + \beta + \gamma = 180^\circ$ , we can reduce $A$ to

$A = 1+ \cos\alpha\cos\beta\cos\gamma$

Let $y = 2\cos\alpha\cos\beta\cos\gamma \quad \cdots \text{Eq 2} \\$ $\Rightarrow y = [\cos(\alpha - \beta) + \cos(\alpha + \beta)]\cos\gamma = [\cos(\alpha - \beta) - \cos\gamma]\cos\gamma \\$ $\Rightarrow \boxed{\cos^2\gamma - \cos(\alpha - \beta)\cos\gamma + y = 0} \\$ This is quadratic equation in $\cos\gamma$ . For $\cos\gamma$ to be real , discriminant should be greater than or equal to 0 $\quad i.e. \\$ $\cos^2(\alpha - \beta) - 4y \geq 0 \\$ $\Rightarrow \large y \leq \frac{\cos^2(\alpha - \beta)}{4} \leq \frac{1}{4}$

Now from Eq 2 we have $2(A - 1) \leq \frac{1}{4}$ $\\ \Rightarrow A \leq \frac{9}{8}$

From Eq 1 we have ,

$\frac{S}{3} \leq \sqrt{\frac{A}{3}} \leq \sqrt{\frac{9/8}{3}} = \frac{\sqrt{6}}{4} \\ \Rightarrow S \leq \frac{3\sqrt{6}}{4}$

So maximum value of $S$ is $\boxed{\frac{3\sqrt{6}}{4}}$ . Comparing with $\frac{a\sqrt{b}}{c}$ we get

$a = 3 , b = 6 , c = 4$

So $a + b + c = \boxed{13}$