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Algebra Difficulty 5.4 AIME, harder Prove it China

Let n≥2n \ge 2 be an integer. Find the largest real number λ\lambda such that the inequality
an2≥λ(a1+a2+⋯+an−1)+2an a_n^2 \ge \lambda(a_1 + a_2 + \cdots + a_{n-1}) + 2a_n
holds for any positive integers a1,a2,…,ana_1, a_2, \dots, a_n satisfying a1<a2<⋯<ana_1 < a_2 < \dots < a_n.

Solution

For ai=ia_i = i, i=1,2,…,ni = 1, 2, \dots, n, we have λ≤(n−2)÷n−12=2n−4n−1\lambda \le (n-2) \div \frac{n-1}{2} = \frac{2n-4}{n-1}.

Since ak≤an−(n−k)a_k \le a_n - (n-k), k=1,2,…,n−1k = 1, 2, \dots, n-1, an≥na_n \ge n, then we have
2n−4n−1∑i=1n−1ai≤2n−4n−1((n−1)an−n(n−1)2)=(2n−4)an−n(n−2)=(n−2)(2an−n)≤(an−2)an. \begin{aligned} \frac{2n-4}{n-1} \sum_{i=1}^{n-1} a_i &\le \frac{2n-4}{n-1} \left( (n-1)a_n - \frac{n(n-1)}{2} \right) \\ &= (2n-4)a_n - n(n-2) \\ &= (n-2)(2a_n - n) \le (a_n - 2)a_n. \end{aligned}

That is, an2≥2n−4n−1(a1+a2+⋯+an−1)+2ana_n^2 \ge \frac{2n-4}{n-1}(a_1 + a_2 + \cdots + a_{n-1}) + 2a_n. So the largest value of λ\lambda is 2n−4n−1\frac{2n-4}{n-1}.

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