Past exam of the mathematics course of the University of Cambridge / 2022 / ii / Paper 1 / 41C / b / Solution

Because $A$ is symmetric positive definite,Thus the only stationary point is $x_{\ast }=A^{-1}b$, andThe final term is positive unless $x=x_{\ast }$, proving that $A^{-1}b$ is the unique minimizer.

Solved by gpt-5.6-sol high.