Page 23 - MATINF Nr.2

On a Crux open problem 23

m
2 − p
ñ Ç å ô
m
m
Suffice it to prove that p 2m 1 + 2 ≥ 4 , ∀p ∈ [1, 2] (because 4 ≤ 3).
2p − 1
m
2 − x
ñ Ç å ô
Define the function g : [1, 2] → R as g (x) = x 2m 1 + 2 , ∀x ∈ [1, 2].
2x − 1
Clearly, g is continuous. Moreover, it is differentiable on [1, 2).
Ç å m−1 "Ç å 1−m 2 #
2 − x 2 − x 4x − 7x + 4
0
We have: g (x) = 2mx 2m−1 − 2 , ∀x ∈ [1, 2).
2x − 1 2x − 1 (2x − 1)
So for finding the minimum of g on the interval [1, 2], suffice it to determine the sign of the
Ç å 1−m 2
2 − x 4x − 7x + 4
expression − on the interval [1, 2).
2x − 1 (2x − 1) 2
2
y + 1 Ç 2 − x å 1−m 4x − 7x + 4
Denote 2x − 1 = y. Then y ∈ [1, 3), x = and − =
2 2x − 1 (2x − 1) 2
Ç å 1−m 2
3 − y 2y − 3y + 3
− .
2y 2y 2
Ç å 1−m 2
3 − y 2y − 3y + 3
Define now the function h : [1, 3) → R as h (y) = − , ∀y ∈ [1, 3).
2y 2y 2
3 ñ Ç 2y å m 2 ô
0
We have: h (y) = − (1 − m) − 1 + , ∀y ∈ [1, 3).
2y 2 3 − y y
Ç å m
2y
But the function is strictly increasing on [1, 3) (as composite of increasing) and
3 − y
2
the function is strictly decreasing on [1, 3) and since 1 − m > 0, we deduce that the function
y
Ç å m
2y 2
ϕ (y) = − (1 − m) − 1 + is strictly decreasing on [1, 3).
3 − y y
On the other hand, ϕ (1) = m > 0 and ϕ (3 − 0) = −∞ < 0, ∃!y 0 ∈ (1, 3) such that
ϕ (y 0 ) = 0.

0
Further, ϕ (y) > 0 ∀y ∈ [1, y 0 ) and ϕ (y) < 0 ∀y ∈ (y 0 , 3) ⇒ h (y) > 0 ∀y ∈ [1, y 0 ) and
0
h (y) < 0 ∀y ∈ (y 0 , 3).
2
But h (1) = 0 and h (3 − 0) = − < 0. Since h is strictly increasing on [1, y 0 ] and strictly
3
decreasing on [y 0 , 3), then ∃!y 1 ∈ (y 0 , 3) such that h (y 1 ) = 0.

0
0
We deduce from here that for a certain x 0 ∈ (1, 2) g (x) > 0 ∀x ∈ (1, x 0 ) and g (x) < 0
∀x ∈ (x 0 , 2).
ln 3
m
m
In conclusion, min g = min {g (1) , g (2)} = 4 . Let’s remark that if m = , then 4 = 3
2 ln 2
ln 3
and equality holds at (1, 1, 1) or at (2, 2, 0) and permutations and if m < , then equality
2 ln 2
holds only at (2, 2, 0) and permutations.
Case 2: p > 2.
m
m
m
m
m
We have: (ab) + (bc) + (ca) ≥ (bc) > 4 .
ln 3
Let study now for m > .
2 ln 2