Openly Hard

Algebra Level 5

Determine the largest integer n n such that i = 1 n x i 2 x n i = 1 n 1 x i \sum_{i=1}^n{x_{i}}^{2}\geq x_{n}\sum_{i=1}^{n-1}x_i for all real numbers x 1 , x 2 , x 3 , , x n x_1,\space x_2, \space x_3,\ldots,\space x_n .


The answer is 5.

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2 solutions

Otto Bretscher
Jan 30, 2016

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 x n 2 ) 2 + 5 n 4 x n 2 0 q(x_1,...,x_n)=\sum_{i=1}^{n}x_i^2-\sum_{i=1}^{n-1}x_nx_i=\sum_{i=1}^{n-1}(x_i-\frac{x_n}{2})^2+\frac{5-n}{4}x_n^2\geq 0 if n 5 n\leq 5 . If n 6 n\geq 6 , we can make q q negative by letting x n = 2 x_n=2 and x i = 1 x_i=1 for all i < n i<n . Thus the answer is 5 \boxed{5} .

Andrea Palma
Mar 5, 2016

We may assume each x i > 0 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 ) \sum_{i=1}^n x_i^2 \geq x_n \cdot \left( \sum_{i=1}^{n-1}x_i\right) dividing by x n 2 x_n^2 we get i = 1 n ( x i x n ) 2 i = 1 n 1 ( x i x n ) \sum_{i=1}^n \left( \dfrac{x_i}{x_n} \right)^2 \geq \sum_{i=1}^{n-1} \left( \dfrac{x_i}{x_n}\right) 1 + i = 1 n 1 ( x i x n ) 2 i = 1 n 1 ( x i x n ) 1 + \sum_{i=1}^{n-1} \left( \dfrac{x_i}{x_n} \right)^2 \geq \sum_{i=1}^{n-1} \left( \dfrac{x_i}{x_n}\right) i = 1 n 1 ( y i y i 2 ) 1 \sum_{i=1}^{n-1} \left( y_i - y_i^2 \right) \leq 1 where y i = x i x n y_i = \dfrac{x_i}{x_n} .

Now we have by elementary calculus (vertex of a parabola) f ( y ) = y y 2 1 4 f(y) = y - y^2 \leq \dfrac{1}{4} , and f ( y ) f(y) reaches its maximum 1 4 \dfrac{1}{4} in y = 1 2 y = \dfrac{1}{2} .

So we can be sure that for n 5 n \leq 5

i = 1 n 1 ( y i y i 2 ) i = 1 n 1 y i 2 y i i = 1 n 1 1 4 n 1 4 = 1 \sum_{i=1}^{n-1} \left( y_i - y_i^2 \right) \leq \sum_{i=1}^{n-1} y_i^2 - y_i \leq \sum_{i=1}^{n-1} \dfrac{1}{4} \leq \dfrac{n-1}{4} = 1

But for n = 6 n = 6 taking y i = 1 2 y_i = \dfrac{1}{2} we have i = 1 6 1 1 4 = 5 4 ≰ 1 \sum_{i=1}^{6-1} \dfrac{1}{4} = \dfrac{5}{4} \not\leq 1

So the max n n for which the inequality always holds in n = 5 n = 5 .

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