设a>0 b>0,a+b=1求证1\a+1\b+1\ab>=8

1a+1b+1ab
=(a+b+1)/ab
=2/ab
又
1=a+b≥2根号(ab)
根号(ab)≤1/2
ab≤1/2
故
1a+1b+1ab
=2/ab
≥2 / (1/4)
=8
证毕

1a+1b+1ab=1a+1b+(a+b)ab=2/a+2/b=2*[(a+b)/a+(a+b)/b]=2*[1+b/a+1+a/b]
=2*[2+a/b+b/a]≥2[2+2]=8

1/a+1/b+1/ab
=(a+b+1)/ab
=2(a+b)/ab
因为
(a-b)²>=0
a²+b²>=2ab
a²+b²+2ab>=4ab
2(a+b)²>=8ab
a+b=1=(a+b)²
所以 2(a+b)/ab>=8