Find the maximum volume of a cone that fits inside a sphere of radius 1.
The ideal cone must have its vertex on the surface of the sphere or else a larger cone will be constructible.  Likewise the circumference of the base must be tangent to the sphere.

[asy]
scale(100);
import graph3;
real s = sqrt(3)/2;

draw(shift(0,0,-1/2)*scale(s,s,3/2)*unitcone,rgb(.6,.6,1));
draw(unitcircle);
real x(real t) {return cos(t);}
real y(real t) {return sin(t);}
real z(real t) {return 0;}
draw(graph(x,y,z,-.69,2.0));
[/asy]

Let $d$ denote the distance from the center of the sphere to the center of the base of the cone.

[asy]
scale(100);

draw(unitcircle);

real s = sqrt(3)/2;

pair A=(0,1);
pair B=(-s,-1/2);
pair C=(s,-1/2);
pair D=(0,-1/2);
pair OO = (0,0);

draw(A--B--C--A--D);
draw(B--OO);
label("$d$",.5D,E);
[/asy]

Since the sphere has radius 1, we can use the Pythagorean Theorem to find other values.

[asy]
scale(100);

draw(unitcircle);

real s = sqrt(3)/2;

pair A=(0,1);
pair B=(-s,-1/2);
pair C=(s,-1/2);
pair D=(0,-1/2);
pair OO = (0,0);

draw(A--B--C--A--D);
draw(B--OO);
label("$d$",.5D,E);
label("$1$",.5A,E);
label("$1$",.5B,NW);
label("$r$",.5(B+D),S);


[/asy]

If $r$ is the radius of the base of the cone, then
\[r^2+d^2=1^2,\]and the height of the cone is
\[h=1+d.\]Therefore, the volume of the cone is
\[V=\frac\pi3r^2h=\frac\pi3(1-d^2)(1+d)=\frac\pi3(1-d)(1+d)^2.\]Thus, we want to maximize $(1-d)(1+d)^2$.

We need a constraint between the three factors of this expression, and this expression is a product.  Let's try to apply the AM-GM inequality by noting that
\[(1-d)+\frac{1+d}2+\frac{1+d}2=2.\]Then
\begin{align*}
\left(\frac23\right)^3 &= \left[\frac{(1-d)+\frac{1+d}2+\frac{1+d}2}3\right]^3 \\
&\geq(1-d)\cdot\frac{1+d}2\cdot\frac{1+d}2,
\end{align*}so
\[
(1-d)(1+d)(1+d)\leq4\left(\frac23\right)^3=\frac{32}{27}.
\]and
\[V=\frac\pi3(1-d)(1+d)^2\leq \frac{\pi}3\cdot\frac{32}{27}= \frac{32\pi}{81}.\]The volume is maximized when the AM-GM inequality is an equality.  This occurs when
\[1-d=\frac{1+d}2=\frac{1+d}2\]so $d=\frac13.$  In this case $h=\frac43$ and
\[r=\sqrt{1-d^2}=\sqrt{\frac89}.\]Indeed, in this case
\[V=\frac\pi3r^2h=\frac\pi3\cdot\frac89\cdot\frac43=\boxed{\frac{32\pi}{81}}.\]