Page 26 - MATINF Nr. 3

26 L.M. Giugiuc

2
2
2
2
2 2
hence the inequality a + a + · · · + a + ka a . . . a ≥ n + k is not fullfiled. In conclusion k ≥ 0.
1
n
2
1 2
n
î √ ä
Lemma 1 (Crux problem 4121). Let t ∈ 1, 2 be a fixed real number. We consider the positive
Ä 2 ä
real number a, b, c and d satisfying a + b + c + d = 2 t +1 and ab + bc + cd + da + ac + bd = 6.
t
2
2
Then min (abcd) = t (2 − t ).
ä
î »
Lemma 2 (generalization of Lemma 1). Let t ∈ 1, n be a fixed real number. We
n−2
Ä 2 ä
consider the positive real numbers a 1 , a 2 , . . . , a n satisfying a 1 + a 2 + · · · + a n = n t +1 and
2t
n (n − 1) n−2 2
X t [n−(n−2)t ]
a i a j = . Then min (a 1 a 2 . . . a n ) = .
2 2
1≤i<j≤n
Proof of the main result
Assume first that a 1 + a 2 + · · · + a n ≥ n. Via the third preliminary we get a 1 + a 2 + · · · + a n =
Ä 2 ä î » ä
n t +1 , with t ≥ 1. Set t ∈ 1, n . According to the second preliminary, all a 1 , a 2 , . . . , a n
2t n−2
are strictly positive. Hence via the Lemma 2, get
2 2
2
t n−2 [n − (n − 2) t ] t 2n−4 [n − (n − 2) t ]
2
2 2
a 1 a 2 . . . a n ≥ ⇒ a a . . . a ≥ ⇒
2 1 2 n 4
2 2
2
Å t + 1 ã 2 t 2n−4 [n − (n − 2) t ]
2
2 2
2
2
2
a + a + · · · + a + ka a . . . a ≥ n 2 − n (n − 1) + k · ,
n
1 2
n
1
2
2t 4
with equality for a 1 = a 2 = · · · = a n−1 = t. So
ã 2
t + 1 t [n − (n − 2) t ]
Å 2 2n−4 2 2
n 2 − n (n − 1) + k · ≥ n + k,
2t 4
Å 2 ã 2 Ç 2n−4 2 2 å
t − 1 t [n − (n − 2) t ]
that is n 2 ≥ k 1 − .
2t 4
î ä
2
So if we set t = x, then x ∈ 1, n and the latter writes as
n−2
2
n (x − 1) 2 î 2 n−2 n−1 2 n ó
≥ k 4 − n x + 2n (n − 2) x − (n − 2) x . (∗)
x
Note that via the AM-GM inequality,
ï ã
n
2 n
2 n−2
n x − 2n (n − 2) x n−1 + (n − 2) x ≤ 4, ∀x ∈ 1, .
n − 2
2 n
2 n−2
Moreover, n x − 2n (n − 2) x n−1 + (n − 2) x = 4 if and only if x = 1. Shall prove next that
n î ó ï n ò
2
2 n
2 n−2
2
n (x − 1) ≤ · x 4 − n x + 2n (n − 2) x n−1 − (n − 2) x ∀x ∈ 1, . (∗∗)
n − 2 n − 2
¶ ©
Further, equality holds if and only if x ∈ 1, n . Inequality (∗∗) translates into
n−2
n
ï ò
n n−1
[(n − 2) x − n] (n − 2) x − nx + nx − (n − 2) ≤ 0∀x ∈ 1, .
n − 2