If $a, b$, and $c$ are positive real numbers with $2 a+4 b+8 c=16$, what is the largest possible value of $a b c$ ?
By AM-GM,

$$
\frac{16}{3}=\frac{2 a+4 b+8 c}{3} \geq \sqrt[3]{(2 a) \cdot(4 b) \cdot(8 c)}=4 \sqrt[3]{a b c} .
$$

Rearranging, we get

$$
\sqrt[3]{a b c} \leq \frac{4}{3} \Longleftrightarrow a b c \leq \frac{64}{27}
$$

This maximum is indeed attainable if we set $2 a=4 b=8 c$, since that is the equality case of AM-GM, which means $2 a=4 b=8 c=\frac{16}{3}$, so $a=\frac{8}{3}, b=\frac{4}{3}, c=\frac{2}{3}$. We conclude that the maximum is indeed $\frac{64}{27}$.