Question

已知a b c属于实数，求证a²/b+b²/c+c²/a≥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<=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

Answer 2

a=√b×a/√b≤(b+a²/b)/2
b=√c×b/√c≤(c+b²/c)/2
c=√a×c/√a≤(a+c²/a)/2
三式相加
a+b+c≤(a+b+c)/2+(a²/b+b²/c+c²/a)/2

Answer 3

http://zhidao.baidu.com/question/175774822.html?push=ql