Questions very similar to this one have been asked, but I was wondering if it was possible to solve it without calculus, in how some other variants of this question are.
One such example is finding the largest volume of a cube inscribed in a sphere of radius r, where you can use the fact that such a cube would have a diagonal length equal to the diameter, and this would allow you to use the pythagorean theorem, so on and so forth.
I wanted to know whether or not it you can solve the titular problem in the same way; without calculus, only using geometric properties of the shapes.
Draw the cross-section of the sphere and the cone as described here. Marking the centre and the radii, by Pythagoras $(h-r)^2 + b^2 = r^2$ where $b$ is the length of the cone's radius. We do not need to find $b$, just $b^2$, and so this gives $V = \frac{1}{3} \pi b^2 = \frac{1}{3} \pi (r^2 - (h-r)^2) = \frac{1}{3} \pi h^2 (2r- h)$.
Then to maximise $V = \frac{1}{3} \pi h^2 (2r - h)$:
$$\sqrt[3]{\frac{1}{2}h \cdot \frac{1}{2}h \cdot (2r - h)} ≤ \frac{2r}{3} \Rightarrow \frac{1}{4} h^2 (2r - h) ≤ \frac{8r^3}{27}$$ $$\Rightarrow \frac{1}{3} \pi h^2 (2r - h) ≤\frac{32\pi r^3}{81}$$
and we can check equality is satisfied when $\frac{1}{2}h = \frac{1}{2}h = (2r - h)$ or $h = \frac{4}{3}r$. When this is substituted into $V$, $\frac{1}{3} \pi (\frac{4}{3}r)^2 (2r - \frac{4}{3}r) = \frac{32 \pi r^3}{81}$, which proves that we have achieved the best upper bound (this is not guaranteed).
Note that the decomposition of $h \cdot h \cdot (2r - h) \cdot (2r - h)$ does not work as the volume depends on $r^4$ instead of $r^3$.