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设n是大于1的自然数,a>0。易知a(?)1时,a-1与n-(1+a+…+a~(n-1))总是异号。所以, (a-1)[n-(1+a+…+a~(n-1))]≤0。即(a-1)(n-(1-a~n)/(1-a))≤0。整理,有a(n-a~(n-1))≤n-1。①显然,①式等号成立的充分必要是a=1。如果a_1,a_2,…,a_n是n个正数,在①中令a=(a_1/((a_1+a_2+…+a_n)/n)~(1/(n-1)),则有a_1~(1/(n-1))·(a_2+…+a_n)/(n-1)≤≤((a_1+a_2+…+a_n)/n)~(n~(n-1)),即((a_1+a_2+…+a_n)/n)~n≥≥a_1((a_1+a_2+…+a_n)/(n-1))~(n-1)。②再在①中令a=(a_2/(a_2+…+a_n)/(n-))~(1/(n-2)),重复上述步骤,并结合②,有
Let n be a natural number greater than 1, a>0. When a(?)1 is known, a-1 and n-(1+a+...+a~(n-1)) always have different signs. Therefore, (a-1) [n-(1+a+...+a~(n-1))] ≤ 0. That is, (a-1)(n-(1-a~n)/(1-a))≤0. Finishing, there is a(n-a~(n-1))≤n-1. 1 Obviously, the fullness and necessity of formula 1 is a=1. If a_1, a_2, ..., a_n are n positive numbers, in 1 let a = (a_1/((a_1+a_2+... +a_n)/n) ~ (1/(n-1)), then there is a_1~ (1/(n-1))(a_2+...+a_n)/(n-1)≤((a_1+a_2+...+a_n)/n)~(n~(n-1)), ie (( A_1+a_2+...+a_n)/n)~n≥≥a_1((a_1+a_2+...+a_n)/(n-1))~(n-1).2 Then in 1 let a=(a_2/( A_2+...+a_n)/(n-))~(1/(n-2)), repeat the above steps and combine 2 with