题目内容
设x,y,z∈R,且满足:x2+y2+z2=1,x+2y+3z=
,则x+y+z=________.
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由柯西不等式,得
(x2+y2+z2)(12+22+32)≥(x+2y+3z)2,
∴(x+2y+3z)2≤14,则x+2y+3z≤
,
又x+2y+3z=
,
∴x=
,
因此x=
,y=
,z=
,
于是x+y+z=
.
(x2+y2+z2)(12+22+32)≥(x+2y+3z)2,
∴(x+2y+3z)2≤14,则x+2y+3z≤
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又x+2y+3z=
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∴x=
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因此x=
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于是x+y+z=
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