已知两正数x，y满足x+y=1，求z=（x+1x）（y+1y）的最小值．

Question

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

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)．