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China Girls' Mathematical Olympiad

Algebra Difficulty 5.1 AIME, harder Prove it China

Assume P1,P2,…,PnP_1, P_2, \dots, P_n (n≥2n \ge 2) is an arbitrary permutation of 1,2,…,n1, 2, \dots, n. Prove that
1P1+P2+1P2+P3+⋯+1Pn−2+Pn−1+1Pn−1+Pn>n−1n+2. \frac{1}{P_1 + P_2} + \frac{1}{P_2 + P_3} + \dots + \frac{1}{P_{n-2} + P_{n-1}} + \frac{1}{P_{n-1} + P_n} > \frac{n-1}{n+2}.

Solution

Proof By Cauchy's inequality, we can get
[(P1+P2)+(P2+P3)+⋯+(Pn−1+Pn)]⋅(1P1+P2+1P2+P3+⋯+1Pn−1+Pn)≥(n−1)2. [(P_1 + P_2) + (P_2 + P_3) + \dots + (P_{n-1} + P_n)] \cdot \left(\frac{1}{P_1 + P_2} + \frac{1}{P_2 + P_3} + \dots + \frac{1}{P_{n-1} + P_n}\right) \ge (n-1)^2.
Therefore
1P1+P2+1P2+P3+⋯+1Pn−1+Pn≥(n−1)22(P1+P2+⋯+Pn)−P1−Pn=(n−1)2n(n+1)−P1−Pn≥(n−1)2n(n+1)−1−2=(n−1)2(n−1)(n+2)−1>(n−1)2(n−1)(n+2)=n−1n+2. \begin{aligned} & \frac{1}{P_1 + P_2} + \frac{1}{P_2 + P_3} + \dots + \frac{1}{P_{n-1} + P_n} \\ \ge & \frac{(n-1)^2}{2(P_1 + P_2 + \dots + P_n) - P_1 - P_n} \\ = & \frac{(n-1)^2}{n(n+1) - P_1 - P_n} \\ \ge & \frac{(n-1)^2}{n(n+1) - 1 - 2} \\ = & \frac{(n-1)^2}{(n-1)(n+2) - 1} \\ > & \frac{(n-1)^2}{(n-1)(n+2)} = \frac{n-1}{n+2}. \end{aligned}

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