2f[(x1+x2)/2]-[f(x1)+f(x2)]
=f{[(x1+x2)/2]^2/(x1*x2)}
=f[(x1^2+2x1x2+x2^2)/4x1x2]
=f(x1/4x2+x2/4x1+1/2)
因为x1/4x2+x2/4x1≥2√x1/4x2*x2/4x1=1/2
所以x1/4x2+x2/4x1+1/2≥1
因为f(x)为单调递减函数
所以f(x1/4x2+x2/4x1+1/2)≤f(1)=0
即2f[(x1+x2)/2]-[f(x1)+f(x2)]≤0
1/2[f(x1)+f(x2)]≥f[(x1+x2)/2]
x1≠x2,所以等号取不到,所以
1/2[f(x1)+f(x2)]＞f[(x1+x2)/2]