求证：(a1+a2+…+an)/n>=(a1*a2*…*an)^(1/n)

题目有问题吧,比如a1=-1,a2=-2,a3=3,n=3
(a1+a2+a3)/n=0=0
设g(x)=a1+a2+...+an=k,f(x)=a1a2...an
f(x)最大时,df=mdg
a2a3a4...an=m
a1a3a4...an=m
a1a2a4...an=m
...
a1a2a3...a(n-1)=m
a1+a2+...+an=k
解得a1=a2=...=an=k/n
即a1a2...an