已知a+b+c+d+e=8,a^2+b^2+c^2+d^2+e^2=16,求e的最大值.我的思路是: (1+a)^2+(1+b)^2+(1+c)^2+(1+d)^2+(1+e)^2=5+2(a+b+c+d+e)+(a^2+b^2+c^2+d^2+e^2)=37∵(1+e)^2≥0∴有(1+a)^2+(1+b)^2+(1+c)^2+(1+d)^2≤37
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![已知a+b+c+d+e=8,a^2+b^2+c^2+d^2+e^2=16,求e的最大值.我的思路是: (1+a)^2+(1+b)^2+(1+c)^2+(1+d)^2+(1+e)^2=5+2(a+b+c+d+e)+(a^2+b^2+c^2+d^2+e^2)=37∵(1+e)^2≥0∴有(1+a)^2+(1+b)^2+(1+c)^2+(1+d)^2≤37](/uploads/image/z/5470815-39-5.jpg?t=%E5%B7%B2%E7%9F%A5a%2Bb%2Bc%2Bd%2Be%3D8%2Ca%5E2%2Bb%5E2%2Bc%5E2%2Bd%5E2%2Be%5E2%3D16%2C%E6%B1%82e%E7%9A%84%E6%9C%80%E5%A4%A7%E5%80%BC.%E6%88%91%E7%9A%84%E6%80%9D%E8%B7%AF%E6%98%AF%3A+%EF%BC%881%2Ba%EF%BC%89%5E2%2B%EF%BC%881%2Bb%EF%BC%89%5E2%2B%EF%BC%881%2Bc%EF%BC%89%5E2%2B%EF%BC%881%2Bd%EF%BC%89%5E2%2B%EF%BC%881%2Be%EF%BC%89%5E2%3D5%2B2%28a%2Bb%2Bc%2Bd%2Be%29%2B%28a%5E2%2Bb%5E2%2Bc%5E2%2Bd%5E2%2Be%5E2%29%3D37%E2%88%B5%EF%BC%881%2Be%EF%BC%89%5E2%E2%89%A50%E2%88%B4%E6%9C%89%EF%BC%881%2Ba%EF%BC%89%5E2%2B%EF%BC%881%2Bb%EF%BC%89%5E2%2B%EF%BC%881%2Bc%EF%BC%89%5E2%2B%EF%BC%881%2Bd%EF%BC%89%5E2%E2%89%A437)
已知a+b+c+d+e=8,a^2+b^2+c^2+d^2+e^2=16,求e的最大值.我的思路是: (1+a)^2+(1+b)^2+(1+c)^2+(1+d)^2+(1+e)^2=5+2(a+b+c+d+e)+(a^2+b^2+c^2+d^2+e^2)=37∵(1+e)^2≥0∴有(1+a)^2+(1+b)^2+(1+c)^2+(1+d)^2≤37
已知a+b+c+d+e=8,a^2+b^2+c^2+d^2+e^2=16,求e的最大值.
我的思路是: (1+a)^2+(1+b)^2+(1+c)^2+(1+d)^2+(1+e)^2=5+2(a+b+c+d+e)+(a^2+b^2+c^2+d^2+e^2)=37
∵(1+e)^2≥0
∴有(1+a)^2+(1+b)^2+(1+c)^2+(1+d)^2≤37
得到:4+2(a+b+c+d)+ a^2+b^2+c^2+d^2=4+2(8-e)+(16-e^2)≤37 然后我就挂了
想问一下是哪里错了··
谢谢,答案是16/5
已知a+b+c+d+e=8,a^2+b^2+c^2+d^2+e^2=16,求e的最大值.我的思路是: (1+a)^2+(1+b)^2+(1+c)^2+(1+d)^2+(1+e)^2=5+2(a+b+c+d+e)+(a^2+b^2+c^2+d^2+e^2)=37∵(1+e)^2≥0∴有(1+a)^2+(1+b)^2+(1+c)^2+(1+d)^2≤37
用柯西不等式做,把e移到常数那边
这道题的前提是a,b,c,d,e大于0吧,要不做不了
∵(1+e)^2≥0应该是大于等于1 ,而且你的做法中几个不等式不能同时取等号
用柯西不等式做最好