Find the smallest positive integer such that there exist real numbers with and .
If you come to the conclusion that no such exists, enter 666.
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Applying Cauchy Shwarz inequality S = k = 1 ∑ n x 0 x k ≤ ( n x 0 2 ) k = 1 ∑ n x k 2 = ( n x 0 2 ) ( 1 − x 0 2 ) Applying G M ≤ A M ≤ n × 4 ( x 0 2 + 1 − x 0 2 ) = 4 n > 1 o r n > 4 Hence smallest integral n= 5 For equality x 1 = x 2 = . . . . = x n = 2 n 1 a n d x 0 = 2 1 For n= 5 , we get x 1 = x 2 = . . . . = x n = 1 0 1 a n d x 0 = 2 1