Question

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！谢谢！

Answer 1

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<=[(a+b)/2]^2）
>=(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

Answer 2

f(x)=(log2(x)-1)/(log2(x)+1)；
f(x1)=(log2(x1)-1)/(log2(x1)+1)；
f(2x2)=(log2(2x2)-1)/(log2(2x2)+1)；
代入f(x1)+f(2x2)=1得到(log2(x1)-1)/(log2(x1)+1)+(log2(2x2)-1)/(log2(2x2)+1)=1，
通分化简得到log2(x1)·log2(2x2)-4=log2(x1·x2)
因为x1、x2均大于2，
所以log2(x1)>1,log2(2x2)>2；
所以log2(x1)·log2(2x2)-4>2；
所以log2(x1·x2)>-2.................(1)
因为x1、x2均大于2，
所以log2(x1·x2)>2..................(2)
结合(1)(2)得到log2(x1·x2)的范围为log2(x1·x2)>2。
f(x1·x2)=(log2(x1·x2)-1)/(log2(x1·x2)+1)
=1-2/[log2(x1·x2)+1]..............【分离常数项】
因为log2(x1·x2)>2
所以log2(x1·x2)+1>3
所以0<2/[log2(x1·x2)+1]<2/3
所以-2/3<-2/[log2(x1·x2)+1]<0
所以1/3<1-2/[log2(x1·x2)+1]<1