如果x，y，z∈R，且x+y+z=a(a>0),x^2+y^2+z^2=a^2/2,求证0≤x≤2a/3,0≤y≤2a/3,0≤z≤2a/3.

来源：百度知道编辑：UC知道时间：2024/07/02 11:53:42

帮帮我啦！要有详细的解答啊

x=a-(y+z)
x^2
=[a-(y+z)]^2
=a^2-2a(y+z)+y^2+z^2+2yz ,因为y^2+z^2>=2yz
<=a^2-2a(a-x)+2y^2+2z^2
=a^2-2a^2+2ax+a^2-2x^2
=2ax-2x^2
3x^2<=2ax
(3x-2a)x<=0
0<=x<=2a/3,或x>=2a/3(舍，因为x+y+z>=2a,可是条件x+y+z=a)
0<=x<=2a/3同理0<=y<=2a/3,0<=z<=2a/3

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