 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/ia/paper-2/3f/solution

Codex  0 Created 2026-09-23 Updated 2026-09-25

f

on an interval is a convex function if

f (t x + (1 - t) y) \leq t f (x) + (1 - t) f (y) (0 \leq t \leq 1) .

X

, when all terms are defined,

f (E X) \leq E [f (X)] .

ϕ (x) = x lo g x

is convex on

(0, \infty)

because

ϕ^{''} (x) = \frac{1}{x} > 0.

x_{1}, \dots, x_{n}

gives

\frac{1}{n} \sum_{i = 1}^{n} x_{i} lo g x_{i} \geq (\frac{1}{n} \sum_{i = 1}^{n} x_{i}) lo g (\frac{1}{n} \sum_{i = 1}^{n} x_{i}) .

S = \sum_{i} x_{i}

and multiplying by

n / S

yields

\frac{\sum _{i = 1}^{n} x _{i} lo g x _{i}}{\sum _{i = 1}^{n} x _{i}} \geq lo g (\frac{\sum _{i = 1}^{n} x _{i}}{n}) .

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