 Solution

ID: 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,

\prod_{i} x_{i}^{a_{i}} = \frac{\prod _{i} y _{i}^{a_{i}}}{G ^{\sum_{i} a_{i}}} = 1.

Part (i) implies

\sum_{i} a_{i} y_{i} / G \geq 1

i \prod y_{i}^{a_{i}} \leq i \sum a_{i} y_{i} .

Equality holds exactly when

x_{i} = 1

for every

i

, or equivalently when all the

y_{i}

are equal.

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