函数f(x)=(log2(x)-1)/(log2(x)+1),若f(x1)+f(2x2)=1 (其中x1、x2均大于2),则f(x1*x2)的最小值为?

题目

函数f(x)=(log2(x)-1)/(log2(x)+1),若f(x1)+f(2x2)=1 (其中x1、x2均大于2),则f(x1*x2)的最小值为?
请不要复制网上的答案,是错的!

答案

化简f(x)=1-2/(log2(x)+1)
f(x1)+f(2x2)
=2-2/(log2(x1)+1)-2/(log2(2x2)+1)
=2-2/(log2(x1)+1)-2/(log2(x2)+2)
=2-2*(log2(x1)+log2(x2)+3)/[(log2(x1)+1)(log2(x2)+2)]=1
故
1/2=(log2(x1)+log2(x2)+3)/【(log2(x1)+1)(log2(x2)+2)】
(对分母用基本不等式ab=(log2(x1)+log2(x2)+3)/【(log2(x1)+log2(x2)+3)/2】^2
=4/(log2(x1)+log2(x2)+3)
解得log2(x1)+log2(x2)>=5
当且仅当log2(x1)+1=log2(x2)+2,即x1=2x2时取等号
f(x1x2)=1-2/(log(x1x2)+1)>=1-2/(5+1)=2/3