Let the sphere have centre $O$ and radius $R$. Let the cone inscribed in the sphere have height $h$ and base radius $r$.

Let the distance of the base of the cone from the centre $O$ be $x$. Then the height of the cone is $h = R + x$.

The radius of the cone's base is given by $r^2 = R^2 - x^2$.

Volume of the cone:

$$V = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi (R^2 - x^2)(R + x)$$ $$V = \frac{1}{3}\pi (R + x)(R - x)(R + x) = \frac{1}{3}\pi (R + x)^2 (R - x)$$

Let us substitute $R + x = y$, where $y \in [R, 2R]$. Then $R - x = 2R - y$.

$$V = \frac{1}{3}\pi y^2 (2R - y) = \frac{1}{3}\pi (2Ry^2 - y^3)$$

Differentiating with respect to $y$:

$$\frac{dV}{dy} = \frac{1}{3}\pi (4Ry - 3y^2)$$

For maximum, $\frac{dV}{dy} = 0$:

$$\frac{1}{3}\pi (4Ry - 3y^2) = 0$$ $$y(4R - 3y) = 0$$ $$y = \frac{4R}{3} \quad (\text{since } y \neq 0)$$

Second derivative test:

$$\frac{d^2V}{dy^2} = \frac{1}{3}\pi (4R - 6y)$$

At $y = \frac{4R}{3}$:

$$\frac{d^2V}{dy^2} = \frac{1}{3}\pi (4R - 8R) = -\frac{4}{3}\pi R < 0$$

Hence, $V$ is maximum when $y = \frac{4R}{3}$, i.e., $h = \frac{4R}{3}$.

Maximum Volume:

At $h = \frac{4R}{3}$:

$$r^2 = R^2 - (h - R)^2 = R^2 - \frac{R^2}{9} = \frac{8R^2}{9}$$ $$V_{\max} = \frac{1}{3}\pi \left(\frac{8R^2}{9}\right)\left(\frac{4R}{3}\right) = \frac{32\pi R^3}{81}$$

Volume of sphere $= \frac{4}{3}\pi R^3$

Ratio:

$$\frac{V_{\max}}{V_{\text{sphere}}} = \frac{32\pi R^3}{81} \div \frac{4\pi R^3}{3} = \frac{8}{27}$$

∴ Hence proved that the volume of the largest cone inscribed in a sphere is $\frac{8}{27}$ of the volume of the sphere.