Past exam of the mathematics course of the University of Cambridge / 2007 / ia / Paper 1 / 7C / ii / Solution

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.