證明:bc/a+ac/b+ab/c
=abc/a²+abc/b²+abc/c²
=abc(1/a²+1/b²+1/c²)
(1/a-1/b)²≥0
((1/a²)+(1/b²)≥2/ab ①
(1/b-1/c)²≥0
(1/b²)+(1/c²)≥2/bc ②
(1/a-1/b)²≥0
(1/a²)+(1/c²)≥2/ac ③
①+②+③
=2/a²+2/b²+2/c²≥2/ab+2/bc+2/ca
=1/a²+1/b²+1/c²≥1/ab+1/bc+1/ca
bc/a+ac/b+ab/c≥abc(1/ab+1/bc+1/ca)=a+b+c
bc/a+ac/b+ab/c≥a+b+c
已知a,b,c均為正數(shù),求證bc/a+ac/b+ab/c大于等于a+b+c
已知a,b,c均為正數(shù),求證bc/a+ac/b+ab/c大于等于a+b+c
數(shù)學(xué)人氣:734 ℃時(shí)間:2020-03-20 08:48:05
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