<s>[INST] Prove this generalization of Cauchy-Schwarz: $|(a.b)c+(b.c)a-(c.a)b|\le|a||b||c|$. If you take $c$ unit perpendicular to $a,b$, Cauchy-Schwarz follows.
I can prove the inequality in a roundabout way. First normalize to unit vectors. Then show that equality holds for 2-dimensional vectors (a bit messy). Then split $c=d+e$ with $d\in\mathrm{Span}(a,b)$, $e\perp d$ and apply $$|(a.b)d+(b.d)a-(a.d)b+(a.b)e|^2=|d|^2+(a.b)^2|e|^2\le|d|^2+|e|^2=1$$
But it's begging for a straightforward proof.
 [/INST] Consider the Gramian matrix for the vectors $a, b, c$:
$$G = \begin{pmatrix} \|a\|^2 & a \cdot b & a \cdot c \\ b \cdot a & \|b\|^2 & b \cdot c \\ c \cdot a & c \cdot b & \|c\|^2 \end{pmatrix}.$$
Recall that $G$ is positive-semidefinite, and positive-definite if and only if $a, b, c$ are linearly independent. Either way, we have $\det G \ge 0$, hence
\begin{align*}
0 &\le \|a\|^2(\|b\|^2\|c\|^2 - (b\cdot c)^2) - (a \cdot b)(\|c\|^2(b \cdot a) - (b \cdot c)(c \cdot a)) \\
&+ (a \cdot c)((b \cdot a)(c \cdot b) - \|b\|^2(c \cdot a)) \\
&= \|a\|^2\|b\|^2\|c\|^2 - \|a\|^2(b \cdot c)^2 - \|b\|^2(a \cdot c)^2 - \|c\|^2(a \cdot b)^2 + 2(a \cdot b)(b \cdot c)(a \cdot c) \\
&= \|a\|^2\|b\|^2\|c\|^2 - \|(a \cdot b)c + (b \cdot c)a - (c \cdot a)b\|^2.
\end{align*}
Therefore, the inequality holds, and equality holds if and only if $a, b, c$ are linearly dependent.
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