Question

已知啊，b,c.均为正数。求证：bc/a+ac/b+ab/c>a+b+c.

Answer 1

a,b,c∈r+
由基本不等式x^2+y^2≥2xy
(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

Answer 2

证：
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
当a，b，c为互不相等的正实数时，bc/a+ac/b+ab/c>a+b+c
如果你想要证的是bc/a+ac/b+ab/c>a+b+c，还需要补充“a，b，c互不相等”这个条件。