 Solution

ID: past-exam-of-the-graph-theory-and-probabilistic-combinatorics-course-of-the-university-of-oxford/2008/b/5/a/ii/solution

x - lo g (1 + x)

x / (1 + x)

, is zero at zero, and is nonnegative for every

x > - 1

. Therefore

lo g (1 + x) \leq x

. Multiplying by

m > 0

and exponentiating gives

(1 + x)^{m} \leq e^{x m} .

Equality occurs at

x = 0

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