Past exam of the mathematics course of the University of Cambridge / 2015 / ia / Paper 1 / 5B (Vectors and Matrices)

The Cauchy-Schwarz inequality in ${\mathbb{R}}^n$ isHere the Euclidean norm is induced by the dot product. If $b=0$, the result is immediate. If $b\ne 0$, the squared Euclidean norm of the component of $a$ perpendicular to $b$ is nonnegative:Rearrangement proves the Cauchy-Schwarz inequality. Equality holds precisely when $a$ and $b$ are linearly dependent, including the cases of a zero vector.

Expanding the squared Euclidean norm and using the Cauchy-Schwarz inequality givesTaking nonnegative square roots gives the triangle inequality. Applying the triangle inequality once more gives

Subtract the two constraints to obtain $x\cdot (a-b)=A-B$. The Cauchy-Schwarz inequality gives $|A-B|\le \Vert x\Vert \Vert a-b\Vert$. Since $a\ne b$, division and squaring give the distance bound from two hyperplane constraintsIf $a=b$ and $A\ne B$, the intersection is empty. If $a=b\ne 0$ and $A=B$, the constraints describe the same hyperplane. Its points satisfy $\Vert x{\Vert }^2\ge A^2/\Vert a{\Vert }^2$, with equality at $x=Aa/\Vert a{\Vert }^2$. The fraction involving $a-b$ is undefined in this case and cannot be used.

If $a=b=0$, both equations are satisfied by every $x$ when $A=B=0$, and otherwise there is no solution. These degenerate cases distinguish an inconsistent constraint from a repeated one.

Apply the triangle inequality to the path from $x_1$ through $y_2$ to $y_1$:This is one direction of the reverse triangle inequality. A second application of the triangle inequality, this time through $x_2$, givesThe resulting four-point triangle inequality follows without any assumptions on coincidences among the four points.