Question

Show that for any vectors a, b in Euclidean space,
$|a\times b{|}^2\le \frac{3\sqrt{3}}{8}|a{|}^2|b{|}^2|a-b{|}^2$.
Remark: Here $\times$ denotes that vector product

Answer 1

In the cases $a=\overrightarrow{0},b=\overrightarrow{0}$, and $a\parallel b$ the inequality is trivial. Otherwise, let us consider a triangle ABC such that $\overrightarrow{CB}=a$ and $\overrightarrow{CA}=b$. From this point on we shall refer to $\alpha ,\beta ,\gamma$ as angles of ABC. Since $|a\times b|=|a||b|sin\gamma$, our inequality reduces to $|a||b|si{n}^3\gamma \le 3\sqrt{3}|c{|}^2/8$, which is further reduced to
$sin\alpha sin\beta sin\gamma \le \frac{3\sqrt{3}}{8}$
using the sine law. The last inequality follows immediately from Jensen's inequality applied to the function $f\left(x\right)=lnsinx$, which is concave for $0<x<\pi$ because $f^{\prime }\left(x\right)=cotx$ is strictly decreasing.
[Jensen's inequality: If $f:I\to R$ is a convex function, then the inequality
$f(a_1{x}_1+....+a_n{x}_n)\le a_n{x}_n)\le a_{1}f(x_1)+....+a_{n}f(x_n)$
holds for all $a_i\ge 0,a_1+....+a_n=1$, and $x_iϵI$. for a concave function the opposite inequality holds.]