对x1,x2∈R，若f(x)=2^x,则[f(x1)+f(x2)]/2与f[(x1+x2)/2]的大小关系是？

(2^x1+2^x2)/2-2^((x1+x2)/2)
>=2*2^((x1+x2)/2-2^((x1+x2)/2)
=0

所以[f(x1)+f(x2)]/2大于等于f[(x1+x2)/2]

[f(x1)+f(x2)]/2>=f[(x1+x2)/2] [x1=x2，时取等号]
画个图吧。
在x轴上取两点P(x1,0),Q(x2,0),
过P,Q点做x轴的垂线，交f(x)于点A,B，线段AB的中点的横坐标为m
[f(x1)+f(x2)]/2=m
过PQ的中点做x轴的垂线交f(x)于点N，其横坐标为n
f[(x1+x2)/2]=n

显然m>n.

1/2[f(x1）+f(x2)]-f[(x1+x2)/2]
=1/2(2^x1+2^x2)-2^[(x1+x2)/2]
=1/2{2^x1+2^x2-2*2^[(x1+x2)/2]}
=1/2{[2^(x1/2)]^2-2*[2^(x1/2)][2^(x2/2)]+[2^(x2/2)]}
=1/2[2^(x1/2)-2^(x2/2)]^2

若x1=x2
则1/2[2^(x1/2)-2^(x2/2)]^2=0，[f(x1)+f(x2)]/2=f[(x1+x2)/2]
若x1≠x2,
则1/2[2^(x1/2)-2^(x2/2)]^2>0，[f(x1)+f(x2)]/2>f[(x1+x2)/2]

所以[f(x1)+f(x2)]/2≥f[(x1+x2)/2]
当x1=x2时取等号