数学题目高一的

来源:百度知道 编辑:UC知道 时间:2024/07/02 11:04:51
a>0,b>0且a+b=1,求:(a+1/a)(b+1/b)≥25/4

0<ab<=1/2(a+b)^2=1/4
(a+1/a)(b+1/b)=ab+1/ab+a/b+b/a
ab+1/ab>=1/4+4=17/4函数在(0,1)上是减函数
a/b+b/a>=2
(a+1/a)(b+1/b)>=25/4

ab+a/b+1/ab+b/a
=(a^2b^2+a^2+1+b^2)/ab
=[a^2b^2+(1-2ab)+1]/ab
=[(ab-1)^2+1]/ab

a+b=1
ab<=(a+b/2)^2=1/4

所以:(ab-1)^2+1≥25/16,0<ab≤1/4,
左式≥25/4.

得证

:(a+1/a)(b+1/b)≥25/4

很简单嘛
吧1 用a +b 替换