已知两正数x，y满足x+y=1，则z＝(x+1x)(y+1y)的最小值为．

Question

已知两正数x，y满足x+y=1，则z＝(x+(1)/(x))(y+(1)/(y))的最小值为______．

Answer 1

z＝(x+(1)/(x))(y+(1)/(y))=xy+(1)/(xy)+(y)/(x)+(x)/(y)=xy+(1)/(xy)+((x+y)2−2xy)/(xy)=xy+(2)/(xy)-2，
令t=xy，则0＜t=xy≤((x+y)/(2))2＝(1)/(4)，（当且仅当x=y时取等号）．
由f（t）=t+(2)/(t)在（0，(1)/(4)]上单调递减，故当t=(1)/(4)时，f（t）=t+(2)/(t)有最小值(33)/(4)，从而
当且仅当x=y=(1)/(2)时，z有最小值为(25)/(4)．
故答案为：(25)/(4)