Maths Olympiad Prep

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, 2015

Algebra Difficulty 7.7 National Olympiad, round 2 Prove it Baltic Way

Let a1,…,ana_1, \dots, a_n be real numbers, fulfilling 0≤ai≤10 \le a_i \le 1 for i=1,…,ni = 1, \dots, n. Prove the inequality
(1−a1n)(1−a2n)⋯(1−an−1n)≤(1−a1a2⋯an)n. (1 - a_1^n)(1 - a_2^n) \cdots (1 - a_{n-1}^n) \le (1 - a_1 a_2 \cdots a_n)^n.

Solution

(1−a1n)(1−a2n)⋯(1−ann)≤((1−a1n)+(1−a2n)+⋯+(1−ann)n)n=(1−a1n+⋯+annn)n. \begin{aligned} (1 - a_1^n)(1 - a_2^n) \cdots (1 - a_n^n) &\le \left( \frac{(1 - a_1^n) + (1 - a_2^n) + \cdots + (1 - a_n^n)}{n} \right)^n \\ &= \left( 1 - \frac{a_1^n + \cdots + a_n^n}{n} \right)^n. \end{aligned}
By applying AM-GM again we obtain
a1a2⋯an≤a1n+⋯+annn⇒(1−a1n+⋯+annn)n≤(1−a1a2⋯an)n, a_1 a_2 \cdots a_n \le \frac{a_1^n + \cdots + a_n^n}{n} \Rightarrow \left(1 - \frac{a_1^n + \cdots + a_n^n}{n}\right)^n \le (1 - a_1 a_2 \cdots a_n)^n,
and hence the desired inequality. □

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