赞 panwangsheng 幼苗
共回答了15个问题采纳率：93.3% 举报
如果a>b>0
则
(a²+b²)/2-ab
=(a²-2ab+b²)/2
=(a-b)²/2
又a>b
故
(a²+b²)/2-ab>0
(a²+b²)/2>ab
成立
但是
若取a=0,b=1
左=ab=0,右=(a²+b²)/2=0.5
满足 左b>0
故,a>b>0是a
则
(a²+b²)/2-ab
=(a²-2ab+b²)/2
=(a-b)²/2
又a>b
故
(a²+b²)/2-ab>0
(a²+b²)/2>ab
成立
但是
若取a=0,b=1
左=ab=0,右=(a²+b²)/2=0.5
满足 左b>0
故,a>b>0是a
1年前
7
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