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题目设x,y,z∈R,且满足x~2+y~2+z~2=1,x+2y+3z=(14)~(1/2),则x+y+z=__.本题言简意赅,内涵朴实、解法多样,思想鲜活,是一道难得一见的好题,下面提供6种解法,供同行参考.解法1(柯西不等式法)由柯西不等式得:(1~2+2~2+3~2)(x~2+y~2+z~2)≥(x+2y+3z)~2即14×(x~2+y~2+z~2)≥14,即x~2+y~2+z~2≥1,当且仅当x/1=y/2=z/3时取等号,又x~2+y~2+z~2=1,故x/1=y/2=z/3,设x/1=y/2=z/3=t,则x=t,y=2t,z
The subject set x, y, z∈R, and satisfy x ~ 2 + y ~ 2 + z ~ 2 = 1, x + 2y + 3z = (14) This title is concise, simple, diverse, fresh thinking, is a rare good question, the following provides six kinds of solutions for peer reference. Solution 1 (Cauchy inequality method) Cauchy inequality: (1 ~ 2 + 2 ~ 2 + 3 ~ 2) (x ~ 2 + y ~ 2 + z ~ 2) ≥ (x + 2y + 3z) ~ 2 14, that is, x ~ 2 + y ~ 2 + z ~ 2≥1 if and only if x / 1 = y / 2 = z / 1, so x / 1 = y / 2 = z / 3 and x / 1 = y / 2 = z / 3 = t, then x = t, y = 2t, z