Question

已知函数f(x)=2^x.x1x2是任意实数且x1不等于x2,证明1/2f(x1)+f(x2)>f[(x1+x2)/2]

Answer 1

x1不等于x2,所以1/2[f(x1)+f(x2)]=1/2[2^x1+2^x2]>1/2*2根号下2^x1*2^x2=根号2^x1*2^x2
=2^[(x1+x2）/2]=f[(x1+x2)/2]

Answer 2

不等式左边=[2^x1+2^x2]/2>2根号(2^x1*2^x2)/2=根号2^(x1+x2){因为x1不等于x2,所以等号取不到}
不等式右边=2^[(x1+x2)/2]=根号2^(x1+x2)得证

Answer 3

函数对x求二阶导数，得f''(x)=ln2*ln2*2^x>0,则得出原函数为凹函数，要证明不等式为凹函数定义式，得证。