National and Regional Contests Moldova Contests Moldova Team Selection Test 2004 Moldova Team Selection Test 5 f(x_1,...,x_n)=\frac{1}{x_1}+\frac{1}{2x_2}+\ldots+\frac{1}{(n-1)x_{n-1}}+\frac{ Problem Statement Let n ∈ N n\in\mathbb{N} n ∈ N A = { ( x 1 , x 2 . . . , x n ) ∣ x i ∈ R + , i = 1 , 2 , . . . , n } A=\{(x_1,x_2...,x_n)|x_i\in\mathbb{R}_{+}, i=1,2,...,n\} A = {( x 1 , x 2 ... , x n ) ∣ x i ∈ R + , i = 1 , 2 , ... , n } f : A → R , f ( x 1 , . . . , x n ) = 1 x 1 + 1 2 x 2 + … + 1 ( n − 1 ) x n − 1 + 1 n x n . f:A\rightarrow\mathbb{R}, f(x_1,...,x_n)=\frac{1}{x_1}+\frac{1}{2x_2}+\ldots+\frac{1}{(n-1)x_{n-1}}+\frac{1}{nx_n}. f : A → R , f ( x 1 , ... , x n ) = x 1 1 + 2 x 2 1 + … + ( n − 1 ) x n − 1 1 + n x n 1 . f ( ( n 1 ) , ( n 2 ) , . . . , ( n n − 1 ) , ( n n ) ) = f ( 2 n − 1 , 2 n − 2 , . . . , 2 , 1 ) . f(\textstyle\binom{n}{1},\binom{n}{2},...,\binom{n}{n-1},\binom{n}{n})=f(2^{n-1},2^{n-2},...,2,1). f ( ( 1 n ) , ( 2 n ) , ... , ( n − 1 n ) , ( n n ) ) = f ( 2 n − 1 , 2 n − 2 , ... , 2 , 1 ) .