Home Competitions RU-SPB 2014 Problem 104
a 1 ≥ a 2 ≥ . . . ≥ a 100 n > 0 a_1\geq a_2\geq... \geq a_{100n}>0 a 1 ≥ a 2 ≥ ... ≥ a 100 n > 0 If we take from ( a 1 , a 2 , . . . , a 100 n ) (a_1,a_2,...,a_{100n}) ( a 1 , a 2 , ... , a 100 n ) some 2 n + 1 2n+1 2 n + 1 numbers b 1 ≥ b 2 ≥ . . . ≥ b 2 n + 1 b_1\geq b_2 \geq ... \geq b_{2n+1} b 1 ≥ b 2 ≥ ... ≥ b 2 n + 1 then b 1 + . . . + b n > b n + 1 + . . . b 2 n + 1 b_1+...+b_n > b_{n+1}+...b_{2n+1} b 1 + ... + b n > b n + 1 + ... b 2 n + 1 Prove, that ( n + 1 ) ( a 1 + . . . + a n ) > a n + 1 + a n + 2 + . . . + a 100 n (n+1)(a_1+...+a_n)>a_{n+1}+a_{n+2}+...+a_{100n} ( n + 1 ) ( a 1 + ... + a n ) > a n + 1 + a n + 2 + ... + a 100 n
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