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