a+b=1,证(a+1/a)(b+1/b)>=25/4

a+b=1,证(a+1/a)(b+1/b)>=25/4
a+b=1,证(a+1/a)(b+1/b)>=25/4

a+b=1,证(a+1/a)(b+1/b)>=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

楼上是个复制党，这个是链接
http://zhidao.baidu.com/question/84058866.html?an=0&si=2
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^...

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楼上是个复制党，这个是链接
http://zhidao.baidu.com/question/84058866.html?an=0&si=2
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(a+1/a)(b+1/b)≥25/4得证
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

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