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(a+1/a)(b+1/b)
=ab+b/a+a/b+1/(ab)
=(a^2b^2+b^2+a^2+1)/(ab)
=[a^2b^2+(a+b)^2-2ab+1]/(ab)
=[a^2b^2+1-2ab+1]/(ab)
=a^2b^2/ab-2ab/ab+2/ab
=ab+2/ab-2
1=a+b>=2√(ab)
所以√(ab)<=1/2
0<ab<=1/4
因为y=x+2/x, 当0<x<√2是减函数
0<ab<=1/4
所以ab+2/ab-2>=(1/4)+2/(1/4)-2=25/4
所以(a+1/a)(b+1/b)>=25/4
=ab+b/a+a/b+1/(ab)
=(a^2b^2+b^2+a^2+1)/(ab)
=[a^2b^2+(a+b)^2-2ab+1]/(ab)
=[a^2b^2+1-2ab+1]/(ab)
=a^2b^2/ab-2ab/ab+2/ab
=ab+2/ab-2
1=a+b>=2√(ab)
所以√(ab)<=1/2
0<ab<=1/4
因为y=x+2/x, 当0<x<√2是减函数
0<ab<=1/4
所以ab+2/ab-2>=(1/4)+2/(1/4)-2=25/4
所以(a+1/a)(b+1/b)>=25/4
参考资料: http://zhidao.baidu.com/question/94249156.html
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