Determine the largest integer n such that i = 1 ∑ n x i 2 ≥ x n i = 1 ∑ n − 1 x i for all real numbers x 1 , x 2 , x 3 , … , x n .
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We may assume each x i > 0 . The inequality can be traslated in the following way. i = 1 ∑ n x i 2 ≥ x n ⋅ ( i = 1 ∑ n − 1 x i ) dividing by x n 2 we get i = 1 ∑ n ( x n x i ) 2 ≥ i = 1 ∑ n − 1 ( x n x i ) 1 + i = 1 ∑ n − 1 ( x n x i ) 2 ≥ i = 1 ∑ n − 1 ( x n x i ) i = 1 ∑ n − 1 ( y i − y i 2 ) ≤ 1 where y i = x n x i .
Now we have by elementary calculus (vertex of a parabola) f ( y ) = y − y 2 ≤ 4 1 , and f ( y ) reaches its maximum 4 1 in y = 2 1 .
So we can be sure that for n ≤ 5
i = 1 ∑ n − 1 ( y i − y i 2 ) ≤ i = 1 ∑ n − 1 y i 2 − y i ≤ i = 1 ∑ n − 1 4 1 ≤ 4 n − 1 = 1
But for n = 6 taking y i = 2 1 we have i = 1 ∑ 6 − 1 4 1 = 4 5 ≤ 1
So the max n for which the inequality always holds in n = 5 .
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Complete the squares: q ( x 1 , . . . , x n ) = ∑ i = 1 n x i 2 − ∑ i = 1 n − 1 x n x i = ∑ i = 1 n − 1 ( x i − 2 x n ) 2 + 4 5 − n x n 2 ≥ 0 if n ≤ 5 . If n ≥ 6 , we can make q negative by letting x n = 2 and x i = 1 for all i < n . Thus the answer is 5 .