Page 27 - MATINF Nr. 3

An unusual configuration 27

d 2
n
But [(n − 2) x − nx n−1 + nx − (n − 2)] = n (x − 1) [(n − 2) x n−3 + · · · + 2x + 1] ≥ 0 on
î dx ó Ä ä
n
1, n , so (n − 2) x − nx n−1 + nx − (n − 2) > 0, ∀x ∈ 1, n . Clearly, (n − 2) x − n < 0,
n−2 n−2
Ä ä î ó
for all x ∈ 1, n , hence (∗∗) is proved. Consider the function f : 1, n → R,
n−2 n−2
n , if x = 1
(
n−2
f (x) = n (x−1) 2 , if 1 < x ≤ n .
2
2 n
2 n−2
x[4−n x +2n(n−2)x n−1 −(n−2) x ] n−2
Ä ä
Clearly, f is continuous, positive, f (1) = f n = n and via (∗∗), f (x) < n , for all
n−2 n−2 n−2
Ä ä Ä ä Ä ä
x ∈ 1, n . So by Weierstrass-Bolzano ∃ 0, n 3 m = min f (x) at some x 0 ∈ 1, n .
n−2 n−2 n−2
x∈[1, n ]
n−2
î ä
As f (x) ≥ m ∀x ∈ 1, n , we get that
n−2
2
n (x − 1) 2 î 2 n−2 n−1 2 n ó ï n ã
≥ m 4 − n x + 2n (n − 2) x − (n − 2) x ∀x ∈ 1, .
x n − 2
ä
» n
î »
n
Consequently, using (∗), if t ∈ 1, , then k max = m. Let t ≥ . Then
n−2 n−2
ã 2
t + 1
Å 2
2 2
2
2
2
2
2
2
2
a + a + · · · + a + ma a . . . a ≥ a + a + · · · + a = n 2 − n (n − 1) .
1 2 n 1 2 n 1 2 n
2t
Ä 2 ä 2 »
But n 2 t +1 − n (n − 1) ≥ n(n−1) , ∀t ≥ n . It remaining to show n(n−1) ≥ n + m ⇐⇒
2t n−2 n−2 n−2
n ≥ m, which is true. We’ll prove that a + a + · · · + a + ma a . . . a ≥ n + m for
2 2
2
2
2
2
n−2 1 2 n 1 2 n
a 1 + a 2 + · · · + a n ≤ −n. But via the substitutions a i → −a i , i = 1, 2, . . . , n this case reduces
to the studied one. For completing, we’ll prove that the admissible domain of k is the interval
[0, m]. Indeed, let k ∈ [0, m]. We have:
h
2 2 2 2 2 2 2 2 2
m a + a + · · · + a + ka a . . . a − (n + k) − k a + a + · · · + a
1 2 n 1 2 n 1 2 n
i
2
2 2
2
2
2
+ma a . . . a − (n + m) = (m − k) a + a + · · · + a − n .
2
1
1 2
n
n
2
2
2
As m − k ≥ 0 and a + a + · · · + a − n ≥ 0, then
1 2 n
2 2 2 2 2 2 2 2 2 2
m a + · · · + a + ka a . . . a − (n + k) − k a + · · · + a + ma a . . . a − (n + m) ≥ 0.
1
n
n
1 2
1
n
n
1 2
2
2
2
2
2 2
But k ≥ 0 and a + a + · · · + a + ma a . . . a − (n + m) ≥ 0 ⇒
1 2 n 1 2 n
2 2 2 2 2 2
m a + a + · · · + a + ka a . . . a − (n + k) ≥ 0
2
1
n
1 2
n
and since m > 0, then
2
2
2
2
2 2
a + a + · · · + a + ka a . . . a ≥ n + k.
1
n
2
1 2
n
This completes the proof of the claim.
√
At the end, we’ll specify that if n = 4 then m = 7 7 − 17. Note that the problem in case
n = 4 was opened by the author in 2016, but no one closed it till date.
Bibliography
https://cms.math.ca/crux/v43/n3/Solutions_43_3.pdf