Given a sphere with radius $$R$$, its volume may be calculated using:

$$V_s = \frac{4}{3} \pi R^3$$

and the volume of a cylinder inscribed in this sphere is given by:

$$V_c = \pi \cdot r^2 \cdot (2h)$$

where $$r^2 = R^2 - h^2$$ due to the Pythagorean theorem.

Now, if we square the volume of the cylinder we find:

$$V_c^2 = (\pi \cdot r^2 \cdot (2h))^2 = 4 \pi^2 \cdot h^2 \cdot (R^2 -h^2) = 2 \pi^2 \cdot (2h^2) \cdot (R^2-h^2) \cdot (R^2-h^2)$$

Using the AM-GM inequality, we find:

$$V_c^2 = 2\pi^2 \cdot (2h^2) \cdot (R^2 - h^2) \cdot (R^2 - h^2) \leq 2\pi^2 \big(\frac{2h^2 + (R^2-h^2) + (R^2-h^2)}{3}\big)$$

which simplifies to:

$$V_c^2 \leq 2\pi^2 \big(\frac{2 \cdot R^2}{3}\big)^3 = \frac{1}{3} \big(\frac{4}{3} \pi R^3\big) = \frac{1}{3} \cdot V_s^2$$

so we may conclude that the volume of the cylinder that may be inscribed in a sphere is given by:

$$V_c \leq \frac{V_s}{\sqrt{3}}$$

QED.