HMMT 二月 2022 · 团队赛 · 第 4 题

4. [30] Suppose n ≥ 3 is a positive integer. Let a < a < · · · < a be an increasing sequence of positive 1 2 n real numbers, and let a = a . Prove that n +1 1 n n X X a a k k +1

. a a k +1 k k =1 k =1 Proposed by: Akash Das Solution 1: We will use induction. The base case is n = 3. In this case, we want to show that a a a a a a 1 2 3 2 3 1

• • • • . a a a a a a 2 3 1 1 2 3 Equivalently, we want to show 2 2 2 2 2 2 2 2 2 a a + a a + a a > a a + a a + a a ⇐⇒ a ( a − a ) + a ( a − a ) > a ( a − a ) 3 1 2 2 3 1 3 2 2 1 3 1 1 2 3 1 2 3 1 3 2 2 2 2 2 ⇐⇒ ( a − a )( a − a ) > ( a − a )( a − a ) 2 1 3 2 3 2 2 1 ⇐⇒ a + a > a + a , 3 2 2 1 which is true. Now assume the claim is true for n ≥ 3. Then, we have that a a a a a a a a 1 3 n n +1 3 4 n +1 1

• • · · · + + > + + · · · + + . a a a a a a a a 3 4 n +1 1 1 3 n n +1 We also have that a a a a a a 1 2 3 2 3 3

• • • • . a a a a a a 2 3 1 1 2 1 Adding the two inequalities and simplifying gives the desired result. 1 1 1 1 Solution 2: The points ( a , ) , ( a , ) , ( a , ) , . . . , ( a , ) form a counter-clockwise oriented poly- 1 2 3 n a a a a 1 2 3 n gon. Thus, we have the area, A , which must be positive, can be calculated by Shoelace theorem: ! n n X X 1 a a k k +1 A = − . 2 a a k +1 k k =1 k =1 Since A is positive, we are done. Solution 3: For 1 ≤ i ≤ n − 1, let r = a /a . Then the inequality becomes i i +1 i 1 1 1 r r · · · r + + · · · + > + r + · · · + r . 1 2 n − 1 1 n − 1 r r r r · · · r 1 n − 1 1 2 n − 1 s − s If we let s = log r and f ( s ) = e − e , this is the same as i i f ( s + · · · + s ) > f ( s ) + · · · + f ( s ) . 1 n − 1 1 n − 1 This follows from the convexity of f and the fact that f (0) = 0.