Question

已知a,b属于正实数,且a+b=1,求y=(a+1/a)(b+1/b)的最小值

Answer 1

(a+1/a)(b+1/b)
=ab+b/a+a/b+1/(ab)
=(a^2b^2+b^2+a^2+1)/(ab)
=[a^2b^2+(a+b)^2-2ab+1]/(ab)
=[a^2b^2+1-2ab+1]/(ab)
=a^2b^2/ab-2ab/ab+2/ab
=ab+2/ab-2

1=a+b>=2√(ab)
所以√(ab)<=1/2
0<ab<=1/4

因为y=x+2/x, 当0<x<√2是减函数
0<ab<=1/4
所以ab+2/ab-2>=(1/4)+2/(1/4)-2=25/4
所以(a+1/a)(b+1/b)>=25/4
最小值为25/4。

Answer 2

解：由a+b=1得b=1-a。故y=(a+1/a)(b+1/b)=y=(a+1/a)[1-a+1/(1-a)]=a+1/a-aa-1+a/(1-a)+1/(a-aa)=(aa-aaa+1-a-aaa+aaaa-a+aa+aa+1)/(a-aa)=(aaaa-2aaa+3aa+aa-2a+2)/(a-aa)
稍后。

Answer 3

y=(a+1/a)(b+1/b)的最小值为4