Past exam of the mathematics course of the University of Cambridge / 2007 / ia / Paper 1 / 7C (Algebra and Geometry)

For the Cauchy-Schwarz inequality, take $\lambda =(\mathbf{x}\cdot \mathbf{y})/(\mathbf{y}\cdot \mathbf{y})$ and use positivity of the inner product:Multiplication by $\mathbf{y}\cdot \mathbf{y}>0$ provesEquality holds exactly when $|\mathbf{x}-\lambda \mathbf{y}{|}^2=0$. Positive definiteness of the inner product makes this equivalent to $\mathbf{x}=\lambda \mathbf{y}$. Conversely, substitution of a scalar multiple gives equality. Thus equality holds precisely for linearly dependent nonzero vectors.

Apply the Cauchy-Schwarz inequality to $(x,y,z)$ and $(1,1,1)$ in the Euclidean inner product on ${\mathbb{R}}^3$. It givesAfter expansion and division by two,The equality criterion from the introductory proof says $x=y=z$. One can also see both the inequality and equality case from $x^2+y^2+z^2-xy-yz-zx=\frac{1}{2}[(x-y{)}^2+(y-z{)}^2+(z-x{)}^2]$.

Set $s=x+y+z$ and $Q=x^2+y^2+z^2$. The expression in the equation is $4(Q+4)-(s+2{)}^2$. Apply the Cauchy-Schwarz inequality in ${\mathbb{R}}^4$ toTheir dot product is $s+2$, and $|\mathbf{e}{|}^2=4$. Thus the expression is nonnegative, and it is zero precisely when equality holds in Cauchy-Schwarz. That requires $\mathbf{v}=\lambda \mathbf{e}$. The fourth coordinate forces $\lambda =2$, and the other three coordinates then giveSubstitution gives $3(4+4+4+4)-2(4+4+4)-4(2+2+2)=48-24-24=0$, so this unique possible choice really is a solution.