由log1/2（x）≥log1/2（2-x）,得:x大于0小于等于1,(*)
f(x)=[log2(x/4)]*[log2(x/2)]=[log2(x)-2]*[log2(x)-1]=[log2(x)]^2-3log2(x)+2=[log2(x)-3/2]^2-3/4
由(*)得log2(x)小于等于0
所以f(x)的最小值=[0-3/2]^2-3/4=2
已知x满足不等式log1/2（x）≥log1/2（2-x）,
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