Classical Inequalities addicts can do this. Part 7

Let $f(\alpha) = \alpha^5$ .

By Cauchy Schwartz Inequality,

$\begin{aligned} (a+b+c)^2 & \le 3(a^2 + b^2 + c^2) \\ (a+b+c)^2 & \le 3(36) \\ a+b+c &\le 6\sqrt{3} \end{aligned}$

Since $f(\alpha)$ is convex at $[0, +\infty)$ , Jensen's inequality can be used.

Then,

$\begin{aligned} 3f\left(\dfrac{a+b+c}{3}\right) \le 3f(2\sqrt{3}) & \le f(a) + f(b) + f(c) \\ a^5 + b^5 + c^5 & \ge 3(2\sqrt{3})^5 = 3*32*9\sqrt{3} = 864\sqrt{3} \end{aligned}$

$\min(a^5 + b^5 + c^5) = 864\sqrt{3} \implies \boxed{s + 2h = 870}$

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Another Sol: Using power-mean inequality,

$\sqrt[5]{\dfrac{a^5 + b^5 + c^5}{3}} \ge \sqrt{\dfrac{a^2 + b^2 + c^2}{3}} \\ a^5 + b^5 + c^5 \ge 3\left( \sqrt{12} \right)^5 = 432\sqrt{12} = 864\sqrt{3} \\ \therefore \boxed{s + 2h = 870}$ .