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Taiwan IMO Selection Camp

Algebra Difficulty 5.3 AIME, harder Prove it Taiwan

Let a1,a2,…,ana_1, a_2, \dots, a_n be nonnegative real numbers such that for every positive integer 1≤k≤n1 \le k \le n,
a1a2⋯ak≥1(2k)! a_1 a_2 \cdots a_k \ge \frac{1}{(2k)!}
Prove that:
a1+a2+⋯+an≥1n+1+1n+2+⋯+12n. a_1 + a_2 + \cdots + a_n \ge \frac{1}{n+1} + \frac{1}{n+2} + \cdots + \frac{1}{2n}.

Solution

We rewrite the left-hand side of the problem as follows:
a1+a2+⋯+an=(1−12)(1⋅2a1)+(13−14)(3⋅4a2)+⋯+(12n−1−12n)((2n−1)⋅2nan)=(1−12−13+14)(1⋅2a1)+(13−14−15+16)(1⋅2a1+3⋅4a2)+⋯+(12n−1−12n)(1⋅2a1+3⋅4a2+⋯+(2n−1)⋅2nan). \begin{aligned} a_1 + a_2 + \cdots + a_n &= \left(1 - \frac{1}{2}\right)(1 \cdot 2a_1) + \left(\frac{1}{3} - \frac{1}{4}\right)(3 \cdot 4a_2) \\ &\quad + \cdots + \left(\frac{1}{2n-1} - \frac{1}{2n}\right)((2n-1) \cdot 2na_n) \\ &= \left(1 - \frac{1}{2} - \frac{1}{3} + \frac{1}{4}\right)(1 \cdot 2a_1) \\ &\quad + \left(\frac{1}{3} - \frac{1}{4} - \frac{1}{5} + \frac{1}{6}\right)(1 \cdot 2a_1 + 3 \cdot 4a_2) + \cdots \\ &\quad + \left(\frac{1}{2n-1} - \frac{1}{2n}\right)(1 \cdot 2a_1 + 3 \cdot 4a_2 + \cdots + (2n-1) \cdot 2na_n). \end{aligned}
By the AM-GM inequality and the condition of the problem, we have:
1⋅2a1≥1, 1⋅2a1+3⋅4a2≥2,⋯ , 1⋅2a1+3⋅4a2+⋯+(2n−1)2nan≥n. 1 \cdot 2a_1 \ge 1, \ 1 \cdot 2a_1 + 3 \cdot 4a_2 \ge 2, \cdots, \ 1 \cdot 2a_1 + 3 \cdot 4a_2 + \cdots + (2n-1)2na_n \ge n.
Hence
a1+a2+⋯+an≥(1−12−13+14)+2(13−14−15+16)+⋯+n(12n−1−12n)=1−12+13−14+15−16+⋯+12n−1−12n=1+12+13+14+15+⋯+12n−1+12n−2(12+14+16+⋯+12n)=1n+1+1n+2+⋯+12n. \begin{aligned} a_1 + a_2 + \cdots + a_n &\ge \left(1 - \frac{1}{2} - \frac{1}{3} + \frac{1}{4}\right) + 2\left(\frac{1}{3} - \frac{1}{4} - \frac{1}{5} + \frac{1}{6}\right) + \cdots \\ &\quad + n\left(\frac{1}{2n-1} - \frac{1}{2n}\right) \\ &= 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \frac{1}{6} + \cdots + \frac{1}{2n-1} - \frac{1}{2n} \\ &= 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \cdots + \frac{1}{2n-1} + \frac{1}{2n} \\ &\quad - 2\left(\frac{1}{2} + \frac{1}{4} + \frac{1}{6} + \cdots + \frac{1}{2n}\right) \\ &= \frac{1}{n+1} + \frac{1}{n+2} + \cdots + \frac{1}{2n}. \end{aligned}

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Source: MathNet, licensed CC-BY-4.0. Statement translated into English from the original; metadata (topic, difficulty) added by this project.