Past exam of the continuous optimisation course of the University of Oxford / 2025 / 2 / b / ii / Solution

Let $G={\prod }_i{y}_i^{a_i}>0$ be the weighted geometric mean and set $x_i=y_i/G$. Since the weights sum to one,Part (i) implies ${\sum }_i{a}_i{y}_i/G\ge 1$, proving the weighted arithmetic-geometric mean inequality:Equality holds exactly when $x_i=1$ for every $i$, or equivalently when all the $y_i$ are equal.