已知a,b,c都是正数,a+b+c=1,求u=(3a^2-a)/(1+a^2)+(3b^2-b)/(1+b^2)+(3c^2-c)/(1+c^2)的最小值,
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![已知a,b,c都是正数,a+b+c=1,求u=(3a^2-a)/(1+a^2)+(3b^2-b)/(1+b^2)+(3c^2-c)/(1+c^2)的最小值,](/uploads/image/z/8649324-36-4.jpg?t=%E5%B7%B2%E7%9F%A5a%2Cb%2Cc%E9%83%BD%E6%98%AF%E6%AD%A3%E6%95%B0%2Ca%2Bb%2Bc%3D1%2C%E6%B1%82u%3D%283a%5E2-a%29%2F%281%2Ba%5E2%29%2B%283b%5E2-b%29%2F%281%2Bb%5E2%29%2B%283c%5E2-c%29%2F%281%2Bc%5E2%29%E7%9A%84%E6%9C%80%E5%B0%8F%E5%80%BC%2C)
已知a,b,c都是正数,a+b+c=1,求u=(3a^2-a)/(1+a^2)+(3b^2-b)/(1+b^2)+(3c^2-c)/(1+c^2)的最小值,
已知a,b,c都是正数,a+b+c=1,求u=(3a^2-a)/(1+a^2)+(3b^2-b)/(1+b^2)+(3c^2-c)/(1+c^2)的最小值,
已知a,b,c都是正数,a+b+c=1,求u=(3a^2-a)/(1+a^2)+(3b^2-b)/(1+b^2)+(3c^2-c)/(1+c^2)的最小值,
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根据式子的对称性
u式中交换a,b的值u式不变
同理交换任何两个量的值 u式均不变
故 取得最小值时
a=b=c=1/3
带入的结果1.8
证明:(3a^2-a)/(1+a^2)>=9a/10-3/10即可
<=>30a^2-10a>=(1+a^2)(9a-3)=9a^3-3a^2+9a-3
<=>9a^3-33a^2+19a-3<=0
<=>(3a-1)(3a^2-10a+3)<=0<=>(3a-1)(3a-1)(a-3)<=0<=>(3a-1)^2(a-3)<=0
显然,(3a-1)^2>=0 ,a...
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证明:(3a^2-a)/(1+a^2)>=9a/10-3/10即可
<=>30a^2-10a>=(1+a^2)(9a-3)=9a^3-3a^2+9a-3
<=>9a^3-33a^2+19a-3<=0
<=>(3a-1)(3a^2-10a+3)<=0<=>(3a-1)(3a-1)(a-3)<=0<=>(3a-1)^2(a-3)<=0
显然,(3a-1)^2>=0 ,a-3<0,故上面不等式成立
同理有
(3b^2-b)/(1+b^2)>=9b/10-3/10
(3c^2-c)/(1+a^2)>=9c/10-3/10
=>u>=9a/10-3/10+9b/10-3/10+9c/10-3/10=0
当且仅当a=b=c=1/3时取最小值
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