已知a>0,b>0a+b=1,求证(a+1⼀a)(b+1⼀b)>=0

2025-05-17 20:37:43
推荐回答(3个)
回答1:

应该是:求证:(a+1/a)(b+1/b)>=25/4

证明:
(a+1/a)(b+1/b)
=ab+a/b+1/ab+b/a
=((ab)^2+a^2+1+b^2)/ab
=[(ab)^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左式≥25/4.

回答2:

(a+1/a)>0
(b+1/b)>0
所以(a+1/a)(b+1/b)>=0

回答3:

不规范的改革和