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已知x+y+z=3,证明X^2+Y^2+Z^2≧3
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已知x+y+z=3,证明X^2+Y^2+Z^2≧3
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答案和解析
根据不等式:a^2+b^2 ≥ 2ab 得:
(x+y+z)^2 = x^2+y^2+z^2+2xy+2yz+2xz ≤ (x^2+y^2+z^2)+(x^2+y^2)+(y^2+z^2)+(x^2+z^2)
即:3(x^2+y^2+z^2) ≥ (x+y+z)^2 =9
∴x^2+y^2+z^2 ≥ 3
即证!
(x+y+z)^2 = x^2+y^2+z^2+2xy+2yz+2xz ≤ (x^2+y^2+z^2)+(x^2+y^2)+(y^2+z^2)+(x^2+z^2)
即:3(x^2+y^2+z^2) ≥ (x+y+z)^2 =9
∴x^2+y^2+z^2 ≥ 3
即证!
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