- 已知a.b正实数，m.n是正整数.m>n.求证…………证明a^m
- 证明a^m+b^m≥a^m-nb^n+a^nb^m-n
- 证明:左-右=(a^m+b^m)-[a^(m-n)b^n+a^nb^(m-n)]
=a^m-a^nb^(m-n)+b^m-a^(m-n)b^n
=a^n*[a^(m-n)-b^(m-n)]+b^n*[b^(m-n)]-a^(m-n)]
=(a^n-b^n)*[a^(m-n)-b^(m-n)]
(1)当a=b时,易证左-右=0
(2)当a0,左-右>0
(3)当a>b时,a^n>b^n.a^n-b^n>0,a^(m-n)>b^(m-n),
a^(m-n)-b^(m-n)>0 ,(a^n-b^n)*[a^(m-n)-b^(m-n)>0,左-右>0
所以 a^m+b^m≥a^m-nb^n+a^nb^m-n