证明:
a,b,c>0
bc/a+ac/b>=2根(bc/a*ac/b)=2c
同理:
ac/b+ab/c>=2a
bc/a+ab/c>=2b
三式相加:
2(bc/a+ac/b+ab/c)>=2(a+b+c)
所以
bc/a+ac/b+ab/c>=a+b+c
a b c 为正实数,求证bc/a+ac/b+ab/c>=a+b+c
a b c 为正实数,求证bc/a+ac/b+ab/c>=a+b+c
数学人气:348 ℃时间:2020-03-30 08:48:06
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