第二数学归纳法
显然当n=2时有
(x1+x2)/2≥(x1x2)^(1/2)
设当n=1,2,...,k时成立,当n=k+1时
则[x1+x2+...+xk+x(k+1)]+(k-1)(x1*x2*...x(k+1)^[1/(1+k)]
=(x1+x2+...+xk)+X(k+1)+(k-1)(x1*x2*...x(k+1)^[1/(1+k)]
≥k（x1*x2*...xk)^(1/k)+k{X(k+1)*[x1*x2*...x(k+1)]^[(k-1)/(k+1)]}(1/k)
≥k(2（x1*x2*...xk)^(1/2k)*{X(k+1)*[x1*x2*...x(k+1)]^[(k-1)/(k+1)]}(1/2k)
=2k(x1*x2*...x(k+1)^[1/(1+k)]
=> (x1+x2+...+xk+X(k+1)≥(k+1)(x1*x2*...x(k+1)^[1/(1+k)]
故当n=k+1时也成立.原命题得证