Question

已知a,b,c属于正实数,求证,（bc/a）+（ac/b）+（ab/c）>=a+b+c

Answer 1

因为a,b,c∈R+

所以：

(bc/2a)+(ac/2b)≥2√[(bc/2a)(ac/2b)]=2√(abc^2/4ab)=c

(bc/2a)+(ab/2c)≥2√[(bc/2a)(ab/2c)]=2√(acb^2/4ac)=b

(ac/2b)+(ab/2c)≥2√[(ac/2b)(ab/2c)]=2√(bca^2/4bc)=a

三式相加即得:

(bc/a)+(ac/b)+(ab/c)≥a+b+c

2)a+b+c=1

由基本不等式：(a+b+c)/3