已知a>0,b>0,c>0,且a、b、c不全相等,求证bc/a+ac/b+ab/c>a+b+c.

由基本不等式x+y≥2√(xy) [x＞0,y＞0,仅当x=y时,x+y=2√(xy)]知：(bc/2a)+(ac/2b)＞2√[(bc/2a)(ac/2b)]=2√(abc²/4ab)=c (bc/2a)+(ab/2c)＞2√[(bc/2a)(ab/2c)]=2√(acb²/4ac)=b (ac/2b)+(ab/2c)＞2√[(...