Question

对于任意的x1，x2∈(0,+∞).若函数f（x）=lgx，试比较[f（x1）+f（x2）]/2与f[（x1+x 2)/2]的大小

请写得详细易懂，谢谢了

Answer 1

1)f(x1)=lgx1
f(x2)=lgx2
2)[f（x1）+f（x2）]/2=(lgx1+lgx2)/2=(lgx1x2)/2
3)x=(x1+x2)/2
f[（x1+x 2)/2]=lg[(x1+x2)/2]
4)[f（x1）+f（x2）]/2-f[（x1+x 2)/2]
=(lgx1x2)/2-lg[(x1+x2)/2]
=lg[(x1x2)^1/2]-lg[(x1+x2)/2]
5)lgx为增函数,所以只需比较
(x1x2)^1/2-(x1+x2)/2<=(x1x2)^1/2-[2(x1x2)^1/2]/2<=0 (x1，x2∈(0,+∞))
6)所以[f（x1）+f（x2）]/2<=f[（x1+x 2)/2]

Answer 2

[f(x1)+f(x2)]/2=(lgx1+lgx2)/2=
[lg(x1*x2)]/2=lg根号下x1*x2

f[(x1+x2)/2]=lg[(x1+x2)/2]

x1,x2属于（0，正无穷）
所以(x1+x2)/2大于等于根号下x1*x2

又因为lgx增函数

所以f[(x1+x2)/2]大于等于[f(x1)+f(x2)]/2