Cauchy-Schwarz inequality

For an orthogonal projection $P$ onto $W$ in an inner product space, apply the Cauchy-Schwarz inequality to $Pu,Pv$ and then use $\Vert Pu\Vert \le \Vert u\Vert$, $\Vert Pv\Vert \le \Vert v\Vert$. If either original vector is zero, equality holds. Otherwise equality holds exactly when both vectors belong to $W$ and are linearly dependent. Necessity follows because equality in the product of the two norm bounds forces equality in each, while equality in Cauchy-Schwarz inequality forces dependence. This retains the zero cases that a division by the original norms would lose.

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 $u+v\ne 0$, the Cauchy-Schwarz inequality gives the unique minimizing third vector $w=-(u+v)/\Vert u+v\Vert$, and the minimum is $u\cdot v-\sqrt{2+2u\cdot v}$. Three coplanar directions separated by $120$ degrees attain the global bound.

If $a\ne b$ and $x\cdot a=A$, $x\cdot b=B$, the Cauchy-Schwarz inequality applied to $x\cdot (a-b)=A-B$ gives $\Vert x\Vert \ge |A-B|/\Vert a-b\Vert$. Equality requires $x$ parallel to $a-b$ and satisfying both original equations. The estimate gives a lower bound on the distance from the origin to the intersection; it need not be sharp if that equality condition is incompatible with the constraints.