Min+Max

Simple algebra shows that, given the constraint that $x + y + z = \tfrac12\pi$ , we have $P(x,y,z) = \cos x\,\sin y \, \cos z \; = \; \tfrac12\big(\cos^2x - \cos^2y + \cos^2z\big)$ and hence, subject to the constraints $x + y + z = \tfrac12\pi$ and $x \ge y \ge z \ge \tfrac{1}{12}\pi$ , $P(x,y,z) \; \le \; P\big(\tfrac12(x+y),\tfrac12(x+y),z\big) \; = \; \tfrac12\cos^2z \; \le \; \tfrac12\cos^2\tfrac{1}{12}\pi \; = \; \tfrac14(\cos\tfrac16\pi + 1) = \frac{\sqrt{3}+2}{8}$ while $P(x,y,z) \; \ge \; P\big(x,\tfrac12(y+z),\tfrac12(y+z)\big) \; = \; P(x,\tfrac14\pi - \tfrac12x,\tfrac14\pi - \tfrac12x\big) \; = \; \tfrac12\cos^2x \; \ge \; \tfrac12\cos^2\tfrac13\pi \; = \; \frac18$ so that $M + m \; = \; \frac{(\sqrt{3} +2) + 1}{8} \; = \; \frac{\sqrt{3}+3}{8}$ making the answer $8+3+3=\boxed{14}$ .