已知a大于0,b大于0,a+b=1,求证(a+1/a)(b+1/b)大于或等于25/4

由均值不等式
a+b≥2√ab
ab≤1/4

证法一
(a+1/a)(b+1/b)
=(a^2+1)/a*(b^2+1)/b
=(a^2b^2+a^2+1+b^2)/ab
=[a^2b^2+(a+b)^2-2ab+1]/ab
=[a^2b^2+(1-2ab)+1]/ab
=[(ab-1)^2+1]/ab
(ab-1)^2+1≥25/16
0<ab≤1/4
(a+1/a)(b+1/b)≥25/4得证
取等号时a=b=1/2

证法二
(a+1/a)(b+1/b)
=ab+1/ab+a/b+b/a
[均值不等式]
≥ab+1/ab+2
[f(x)=x+1/x在x∈(0,1]上单调递减]
≥1/4+1/(1/4)+2
=25/4
取等号时a=b=1/2

(a+1/a)(b+1/b)
=ab+1/ab+a/b+b/a
>或=2+ab+1/ab-------此时取等号是a=b

由基本不等式有
a+b>或=2根号ab,所以ab<或=1/4

设g(x)=x+1/x,求导得在四分之一到正无穷的范围内,取到的最大值是g(1/4)=4+1/4-----------------------此时取等号是a=b

综上,(a+1/a)(b+1/b)大于或等于2+4+1/4=25/4