g(x)=a^x

题目

g(x)=a^x
求证：
g[(x1+x2)/2]小于等于[g(x1)+g(x2)/2]

答案

g(x1)+g(x2)
=a^x1+a^x2
由基本不等式
a^x1+a^x2
≥2√(a^x1*a^x2)
=2a^[(x1+x2)/2]
=2g[(x1+x2)/2]
即[g(x1)+g(x2)]/2≥g[(x1+x2)/2]