= Minimum pairwise dot-product sum of three unit vectors
{title2=$u\cdot v+v\cdot w+w\cdot u\geq-3/2$}
For three <unit vectors>, the squared <Euclidean norm> of their sum is $3+2S$, proving the displayed bound; equality holds exactly when $u+v+w=0$. For fixed unit vectors $u,v$ with $u+v\ne0$, the <Cauchy-Schwarz inequality> gives the unique minimizing third vector $w=-(u+v)/\|u+v\|$, and the minimum is $u\cdot v-\sqrt{2+2u\cdot v}$. Three coplanar directions separated by $120$ degrees attain the global bound.
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