证明: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