Page 22 - MATINF Nr.2

On a Crux open problem

1
Leonard Mihai Giugiuc s , i Ardak Mirzakhmedov 2

In the volume 43, issue 3 of Crux Mathematicorum Magazine, has been featured the following
problem:
Let m be a positive real number. Prove (or disprove) that for all a, b, c ≥ 0 with a + b + c =
ab + bc + ca > 0, the following inequality holds:
m
m
m
m
(ab) + (bc) + (ca) ≥ min {3, 4 } .
(Leonard Giugiuc)
Also, the problem has been promoted in varies specialized communities, but no one answered to
date.

Meanwhile, the author and Ardak Mirzakhmedov, Kazakhstan, found, independently, a
solution based on same principles.
We present below their proof.
ln 3
Ç ô
First, we’ll consider m ∈ 0, ⇒ m < 1.
2 ln 2
2
2
Let a + b + c = ab + bc + ac = k. Since (a + b + c) ≥ 3 (ab + bc + ac) then k ≥ 3k and
since k > 0 then we deduce that k ≥ 3.
b + c − bc
Choose WLOG bc = max {ab, bc, ac}. From above, bc ≥ 1. Also, a = .
b + c − 1
2s − p 2
2
Denote b + c = 2s and bc = p . By AM − GMs ≥ p, from above p ≥ 1 and a = .
2s − 1
2t − p 2
Case 1: 1 ≤ p ≤ 2. For any p thus fixed, consider the function f p (t) = on [p, ∞).
2t − 1
2s − p 2 2p − p 2
2
Since p ≥ 1, then clearly f p is increasing ⇒ f p (s) ≥ f p (p) ⇒ a = ≥ =
2s − 1 2p − 1
2 − p
Ç å
p.
2p − 1
Ç 2 − p å
m
m
m m
By the AM − GM inequality, (ab) + (ca) m ≥ 2a p . But a ≥ p ⇒ (ab) +
2p − 1
Ç å m
m 2 − p 2m
(ca) ≥ 2 p .
2p − 1
m 2m
Since (bc) = p , we get:
m
Ç 2 − p å m ñ Ç 2 − p å ô
m
m
m
(ab) + (bc) + (ca) ≥ p 2m + 2 p 2m = p 2m 1 + 2 , ∀p ∈ [1, 2] .
2p − 1 2p − 1
1
Professor, Colegiul Nat , ional, ,,Traian”, Drobeta Turnu Severin, leonardgiugiuc@yahoo.com
2
Almaty, Kazakhstan, ardak agu@yahoo.com
22