Show that for any vectors $a, b$ in Euclidean space,
\[|a \times b|^3 \leq \frac{3 \sqrt 3}{8} |a|^2 |b|^2 |a-b|^2\]
Remark. Here $\times$ denotes the vector product.
We know that if $\theta$ is the angle between the two vectors a and b, then $|a \times b| =|a||b| \sin \theta$ and $|a-b|^2 = |a|^2 + |b|^2 - 2 |a|b| \cos \theta$.
Hence,
$\frac{3 \sqrt{3}}{8} |a|^2|b|^2|a-b|^2 = \frac{3 \sqrt{3}}{8} |a|^2|b|^2( |a|^2 + |b|^2 - 2 |a|b| \cos \theta )$
$\geq \frac{3 \sqrt{3}}{8} |a|^2|b|^2( 2|a||b| - 2 |a|b| \cos \theta )$
$= |a|^3|b|^3 \frac{3 \sqrt{3}}{4} ( 1 - \cos \theta )$
$\geq |a|^3|b|^3 \sin^3 \theta$
$= |a \times b|^3$.
There is equality when $|a|=|b|$ (due to the AM-GM) and $\theta \in \{ 2\pi n, \frac{\pi}{3} + 2\pi n\}$, for integer $n$ (since $f(x) = \frac{3\sqrt{3}}{4} (1-\cos \theta) - \sin^3 \theta$ has its global minimum at $0$ when $x=2\pi n$ or $\frac{\pi}{3} + 2\pi n$).