已知x,y属于（0,2）,且xy=1,求2/(2-x)+4/(4-y)的最小值

2/(2-x)+4/(4-y)=1/(1-x/2)+1/(1-y/4)因为公式：1/a+1/b≥4/(a+b)故：原式=2/(2-x)+4/(4-y)=1/(1-x/2)+1/(1-y/4)≥4/(2-x/2-y/4)因为xy=1,所以（x/2）·（y/4）=1/8≤1/4*（x/2+y/4）^2→x/2+y/4≥1/√2.所以,有:原...