Eighteenth STARS OF MATHEMATICS Competition

Question

Fix an integer n 2. Determine the least possible value the sum x_2 + x_3 + + x_nx_1 + x_1 + x_2 + + x_nx_2 + + x_1 + x_2 + + x_n-1x_n may achieve, as x_1, x_2, , x_n run through all positive real numbers.

Accepted Answer

The minimum exists, as the summands are all non-negative integers; it is equal to

(n - 1)^{2}

and is achieved if, for instance,

x_{1} = n

and

x_{2} = \dots = x_{n} = n + 1

; the verification is routine.

Let

s = x_{1} + x_{2} + \dots + x_{n}

and let

S

denote the sum in the statement. Note that

⌊ \frac{x _{1} + \dots + x _{k - 1} + x _{k + 1} + \dots + x _{n}}{x _{k}} ⌋ = ⌊ \frac{s - x _{k}}{x _{k}} ⌋ = ⌊ \frac{s}{x _{k}} - 1 ⌋ = ⌊ \frac{s}{x _{k}} ⌋ - 1 > (\frac{s}{x _{k}} - 1) - 1 = \frac{s}{x _{k}} - 2, k = 1, \dots, n .

Sum over

k = 1, 2, \dots, n

to get

S > s \cdot (\frac{1}{x _{1}} + \frac{1}{x _{2}} + \dots + \frac{1}{x _{n}}) - 2 n = = (x_{1} + x_{2} + \dots + x_{n}) (\frac{1}{x _{1}} + \frac{1}{x _{2}} + \dots + \frac{1}{x _{n}}) - 2 n \geq n^{2} - 2 n,

by the AM-HM (or Cauchy-Schwarz or Chebyshev) inequality. Finally, as

S

and

n^{2} - 2 n

are both integers,

S \geq n^{2} - 2 n + 1 = (n - 1)^{2}

, as desired. This ends the proof.