Question

（不等式選講選做題）x、y＞0，x+y=1，則(x+

1

x

)(y+

1

y

)的最小值為_(kāi)_____．

Answer 1

(x+

1

x

)(y+

1

y

)=xy+

x

y

+

y

x

+

1

xy

=

x2y2+x2+y2+1

xy

=

x2y2+(x+y)2-2xy+1

xy

=xy+

2

xy

-2
∵x+y=1
∴xy≤(

x+y

2

)2=

1

4

令xy=t則0＜t≤

1

4

∴(x+

1

x

)(y+

1

y

)=t+

2

t

-2
∵t+

2

t

-2在(0，

1

，4

]遞減
當(dāng)t=

1

4

最小為

25

4

故答案為

25

4