a、b、c都是正实数,求证:(b+c)/2a+(a+c)/2b+(a+b)/2c>=2a/(b+c)+2b/(a+c)+
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a、b、c都是正实数,求证:(b+c)/2a+(a+c)/2b+(a+b)/2c>=2a/(b+c)+2b/(a+c)+2c/(a+b),
![a、b、c都是正实数,求证:(b+c)/2a+(a+c)/2b+(a+b)/2c>=2a/(b+c)+2b/(a+c)+](/uploads/image/z/18078916-4-6.jpg?t=a%E3%80%81b%E3%80%81c%E9%83%BD%E6%98%AF%E6%AD%A3%E5%AE%9E%E6%95%B0%2C%E6%B1%82%E8%AF%81%EF%BC%9A%28b%2Bc%29%2F2a%2B%28a%2Bc%29%2F2b%2B%28a%2Bb%29%2F2c%3E%3D2a%2F%28b%2Bc%29%2B2b%2F%28a%2Bc%29%2B)
(b+c)/2a+(a+c)/2b+(a+b)/2c>=2a/(b+c)+2b/(a+c)+2c/(a+b),
=b/2a+c/2a+a/2b+c/2b+a/2c+b/2c-2a/(b+c)-2b/(a+c)-2c/(a+b)
=b/2a+b/2c-2b/(a+c) + c/2a+c/2b-2c/(a+b) + a/2c+a/2b-2a/(b+c)
b/2a+b/2c-2b/(a+c)=b(1/2a+1/2c-2/(a+c))=b(a-c)(a-c)/(2ac(a+c))
c、c都是正实数(a-c)(a-c)>=0 2ac(a+c)>0 所以 b(a-c)(a-c)/(2ac(a+c))>=0
同理c/2a+c/2b-2c/(a+b)>=0 a/2c+a/2b-2a/(b+c)>=0
(b+c)/2a+(a+c)/2b+(a+b)/2c>=2a/(b+c)+2b/(a+c)+2c/(a+b)
=b/2a+c/2a+a/2b+c/2b+a/2c+b/2c-2a/(b+c)-2b/(a+c)-2c/(a+b)
=b/2a+b/2c-2b/(a+c) + c/2a+c/2b-2c/(a+b) + a/2c+a/2b-2a/(b+c)
b/2a+b/2c-2b/(a+c)=b(1/2a+1/2c-2/(a+c))=b(a-c)(a-c)/(2ac(a+c))
c、c都是正实数(a-c)(a-c)>=0 2ac(a+c)>0 所以 b(a-c)(a-c)/(2ac(a+c))>=0
同理c/2a+c/2b-2c/(a+b)>=0 a/2c+a/2b-2a/(b+c)>=0
(b+c)/2a+(a+c)/2b+(a+b)/2c>=2a/(b+c)+2b/(a+c)+2c/(a+b)
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