已知a>0,b>0,c>0,求证:2/(a+b)+2/(b+c)+2/(c+a)>=9/(a+b+c)这道题怎么解
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![已知a>0,b>0,c>0,求证:2/(a+b)+2/(b+c)+2/(c+a)>=9/(a+b+c)这道题怎么解](/uploads/image/z/5596613-53-3.jpg?t=%E5%B7%B2%E7%9F%A5a%3E0%2Cb%3E0%2Cc%3E0%2C%E6%B1%82%E8%AF%81%3A2%2F%28a%2Bb%29%2B2%2F%28b%2Bc%29%2B2%2F%28c%2Ba%29%3E%3D9%2F%28a%2Bb%2Bc%29%E8%BF%99%E9%81%93%E9%A2%98%E6%80%8E%E4%B9%88%E8%A7%A3)
已知a>0,b>0,c>0,求证:2/(a+b)+2/(b+c)+2/(c+a)>=9/(a+b+c)这道题怎么解
已知a>0,b>0,c>0,求证:2/(a+b)+2/(b+c)+2/(c+a)>=9/(a+b+c)这道题怎么解
已知a>0,b>0,c>0,求证:2/(a+b)+2/(b+c)+2/(c+a)>=9/(a+b+c)这道题怎么解
附证 设2x=a+b,2y=b+c,2z=c+a,则所证不等式等价于
1/x+1/y+1/z>=9/(x+y+z)
(x+y+z)/x+(x+y+z)/y+(x+y+z)/z>=9
y/x+z/x+x/y+z/y+x/z+y/z>=6
(y/x+x/y)+(z/x+x/z)+(y/z+z/y)>=6.
因为 y/x+x/y>2,z/x+x/z>2,y/z+z/y>=2.
所以上式显然成立.
如果了解柯西不等式,那么很简单
(a+b+b+c+c+a)*[1/(a+b)+1/(b+c)+1/(c+a)]>9
2/(a+b)+2/(b+c)+2/(c+a)>9/(a+b+c).