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高中代数第二册中有这样的两个不等式:已知a,b∈R~+,并且a≠b,那么a~3+b~3>a~2b+ab~2;a~5+b~5>a~3b~2+a~2b~3。本文将其推广为更一般的不等式。即下面的 [定理] 设a_1,a_2,…,a_n,m,a,k∈R~+,且m=a+(n-1)k,n≥2,则a_1~m+a_2~m+…+a_n~m≥a_1~a a_2~k…a_n~k+a_1~ka_2~aa_3~k…a_n~k+…a_1~k…a_(n-1)~ka_n~a…(A)成立。(当且仅当a_1=a_2=…=a_n时取“=”号)。证:对n用数学归纳法。①当n=2时,m=a+k,a_1~m+a_2~m-(a_1~aa_2~k+a_1~ka_2~a)=(a_1~a-a_2~a)(a_1~k-a_2~k)≥0,仅当a_1=a_2时取“=”号。命题成立。
There are two inequalities in the second volume of high school algebra: known a, b∈R~+, and a≠b, then a~3+b~3>a~2b+ab~2;a~5+b ~5>a~3b~2+a~2b~3. This article will promote it as a more general inequality. That is, the following [theorems] Let a_1,a_2,...,a_n,m,a,k∈R~+, and m=a+(n-1)k,n≥2, then a_1~m+a_2~m+...+ A_n~m≥a_1~a a_2~k...a_n~k+a_1~ka_2~aa_3~k...a_n~k+...a_1~k...a_(n-1)~ka_n~a...(A) holds. (If and only if a_1=a_2=...=a_n, the “=” sign is taken). Proof: Mathematical induction for n. 1 When n=2, m=a+k, a_1~m+a_2~m-(a_1~aa_2~k+a_1~ka_2~a)=(a_1~a-a_2~a)(a_1~k-a_2) ~k) ≥ 0, only when a_1 = a_2 is taken “=” sign. The proposition is established.