1/(a-b)+1/(b-c)+m/(c-a) 对满足a>b>c恒成立,则m的取值范围1/(a-b)+1/(b-c)+m/(c-a) >0,对满足a>b>c恒成立,则m的取值范围
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![1/(a-b)+1/(b-c)+m/(c-a) 对满足a>b>c恒成立,则m的取值范围1/(a-b)+1/(b-c)+m/(c-a) >0,对满足a>b>c恒成立,则m的取值范围](/uploads/image/z/10882836-36-6.jpg?t=1%2F%28a-b%29%2B1%2F%28b-c%29%2Bm%2F%28c-a%29+%E5%AF%B9%E6%BB%A1%E8%B6%B3a%3Eb%3Ec%E6%81%92%E6%88%90%E7%AB%8B%2C%E5%88%99m%E7%9A%84%E5%8F%96%E5%80%BC%E8%8C%83%E5%9B%B41%2F%28a-b%29%2B1%2F%28b-c%29%2Bm%2F%28c-a%29+%3E0%2C%E5%AF%B9%E6%BB%A1%E8%B6%B3a%3Eb%3Ec%E6%81%92%E6%88%90%E7%AB%8B%2C%E5%88%99m%E7%9A%84%E5%8F%96%E5%80%BC%E8%8C%83%E5%9B%B4)
1/(a-b)+1/(b-c)+m/(c-a) 对满足a>b>c恒成立,则m的取值范围1/(a-b)+1/(b-c)+m/(c-a) >0,对满足a>b>c恒成立,则m的取值范围
1/(a-b)+1/(b-c)+m/(c-a) 对满足a>b>c恒成立,则m的取值范围
1/(a-b)+1/(b-c)+m/(c-a) >0,对满足a>b>c恒成立,则m的取值范围
1/(a-b)+1/(b-c)+m/(c-a) 对满足a>b>c恒成立,则m的取值范围1/(a-b)+1/(b-c)+m/(c-a) >0,对满足a>b>c恒成立,则m的取值范围
1/(A-B)+1/(B-C)+M/(C-A)>0
1/(A-B)+1/(B-C)>M/(A-C)
左边运用柯西不等式推论,等到1/(A-B)+1/(B-C)>=4/(A-C)
因为等式须恒成立,所以M/(A-C)C
所以M