a,b属于正实数,a+b=1,求(1+1/a)(1+1/b)>=9

(1+1/a)(1+1/b)
=[1+(a+b)/a][1+(a+b)/b]
=(1+1+b/a)(1+1+a/b)
=(2+b/a)(2+a/b)
=4+2b/a+2a/b+1
=5+2(b/a+a/b)
≥5+2×2√[(b/a)(a/b)]
=5+2×2
=9
即(1+1/a)(1+1/b)≥9

a+b=1
(a+1/a)(1+1/b)
=(a+1)/a*(b+1)/b
=(ab+a+b+1)/ab
=1+(a+b)/ab+1/ab》9
1+2/ab=9
=>
0<ab<1/4
a+b=1
a=1-b
0<b(1-b)<1/4
=>b(1-b)>0 => 0<b<1
=>b(1-b)<1/4 => -b^2+b-1/4<0 =>(b-1/2)^2>0
所以得
0<b<1,0<a<1，且0<ab<1/4