论文部分内容阅读
本文介绍一个结构简单但应用广泛的不等式.定理设 a>0,b>0,n∈N,则(a~(n+1))/b~n≥(n+1)a-nb(*),当且仅当 a=b 时等号成立.证明:(*)(?)a~(n+1)≥(n+1)ab~n-
This paper introduces a simple structure but widely used inequality. Theorem Let a>0, b>0, n∈N, then (a~(n+1))/b~n≥(n+1)a-nb(* ) If and only if a=b the equal sign holds. Proof: (*)(?)a~(n+1)≥(n+1)ab~n-