設(shè)f(x)=e^(x-1)– x,f’(x)=e^(x-1)-1; f”(x)=e^(x-1)
f(1)=0,f’(1)=0,f”(x)>0,∴f(x)在x=1有絕對的最低值
f(x)=e^(x-1)- x≥f(1)=0
∴e^(x-1) ≥ x--------------------------------------(1)
設(shè)xi>0,i=1,n
設(shè)算術(shù)平均值a=(x1+x2+x3+…+xn)/n,a>0,
從(1),x/a ≤ e^(x/a-1) -------------------(2)
從(2),(x1/a)*(x2/a)*(x3/a)*…*(xn/a ) ≤ e^(x1/a-1) e^(x2/a-1)e^(x3/a-1)… e^(xn/a-1)
=e^(x1/a-1+x2/a-1+x3/a-1+…xn/a-1)=e^[(x1+x2+x3+…+xn)/a-n]
=e^[na/a-n]=e^0=1
∴(x1/a)*(x2/a)*(x3/a)*…*(xn/a )=(x1*x2*x3*…*xn)/a^n ≤ 1
(x1*x2*x3*…*xn) ≤ a^n
(x1*x2*x3*…*xn)^(1/n) ≤ a ,即算術(shù)平均數(shù)大于等于幾何平均數(shù)