a,b,c为正实数,求证：ab/c+bc/a+ac/b>=a+b+c

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<=根号[(a^2+b^2+c^2)/3],等号当且仅当a=b=c时成立
所以根号a+根号b+根号c<=3根号[(a+b+c)/3]=根号3
等号当且仅当a=b=c=1/3时成立