Inscribed Bonanza

Let $V_{s} = \dfrac{4}{3}\pi R^3, \:\ V_{p} = \dfrac{1}{3}\pi r^2 H, \:\ V_{c} = \pi x^2 h$ and $V_{m} = \dfrac{1}{3}y^2 h$ .

$R^2 = (H - r)^2 + r^2 = H^2 - 2HR + R^2 + r^2 \implies r^2 = 2HR - H^2 \implies V_{p} = \dfrac{1}{3}\pi(2H^2R - H^3) \implies$

$\dfrac{dV_{p}}{dH} = \dfrac{\pi}{3}RH(4R - 3H) = 0 \:\ H \neq 0 \implies H = \dfrac{4}{3}R \implies r = \dfrac{2\sqrt{2}}{3}R \implies \boxed{V_{p} = (\dfrac{2}{3})^3 V_{s}}$

The two triangles in the above diagram are similar $\implies \dfrac{r - x}{r} = \dfrac{h}{H} \implies h = \dfrac{(r - x)H}{r} \implies V_{c} = \dfrac{\pi H}{r}(rx^2 - x^3) \implies \dfrac{dV_{c}}{dx} = \dfrac{\pi Hx}{r}(2r - 3x) = 0$ $x \neq 0 \implies x = \dfrac{2}{3}r \implies h = \dfrac{H}{3}$

$r = \dfrac{2\sqrt{2}}{3}R$ and $H = \dfrac{4}{3}R \implies x = \dfrac{4\sqrt{2}}{9}R$ and $h = \dfrac{4}{9}R \implies \boxed{V_{c} = (\dfrac{2}{3})^5V_{s}}$

$y = \sqrt{2}x$ and $x = \dfrac{4\sqrt{2}}{9}R \implies y = \dfrac{8}{9}R \implies \boxed{V_{m} = \dfrac{1}{\pi}(\dfrac{2}{3})^6 V_{s}}$

$\implies \dfrac{V_{p} * V_{c} * V_{m}}{V_{s}^3} = \dfrac{1}{\pi}(\dfrac{2}{3})^{14} =$ $\dfrac{1}{\pi}(\dfrac{16384}{4782969}) = \dfrac{1}{\pi}(\dfrac{a}{b}) \implies a + b = \boxed{4799353}$ .