Question

已知a大于0、b大于0，且a+b=1，求证（1+1/a）（1+1/b）大于等于9

Answer 1

证:  1>a>0,1>b>0

      (1+1/a)(1+1/b)-9=1+1/(ab)+1/a+1/b-9=1/(ab)+1/a+1/b-8=(1+a+b)/(ab)-8=2(a+b)/(ab)-8

      a+b=1=(a+b)(a+b)

Answer 2

(1+1/a)(1+1/b)=(1+(a+b)/a)(1+(a+b)/b)=(2+b/a)*(2+b/a)=5+2(a/b+b/a)>=5+2*2根号（a/b*b/a）=9,且a=b=1/2是取最小值9

Answer 3

=1+1/a+1/b+1/ab=1+(a+b+1)/ab=1+2/ab
=1+2/a(1-a)
由于a（1-a）<1/4
所以，原式大于9

Answer 4

(1+1/a)(1+1b)=(1+(a+b)/a)(1+(a+b)/b)=(2+b/a)(2+a/b)=5+2(a/b+b/a)>=9