Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 339 / 1 / b / iii / Solution

The defining minimization, compared with the candidate $x_k$, shows that $f(x_{k+1})\le f(x_k)$. Put $u=x^{\ast }$ in part ii and sum from $j=0$ to $k-1$. The squared distances telescope, while monotonicity givesHence