Page 30 - RMGO 3

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30 Leonard GIUGIUC and Bogdan SUCEAVA
3
h 2(1+t) (1−3t) i
2
3
Finally, let the function h (x) = x 2 2x − 4 + s 2 − = 2x − 4x −
x x 3
3
2(1+t) (1−3t)
x + s, ∀x > 0. We have:
0
3
3
4
h (x) 3x − 4x + (1 + t) (1 − 3t)
= ∀x > 0.
2 x 2
3
4
3
Shall study now the polynomial Q (x) = 3x − 4x + (1 + t) (1 − 3t) on (0, ∞).
2
0
Q (x) = 12x (x − 1) ∀x > 0, such that Q is strictly decreasing on (0, 1]
3
and strictly increasing on [1, ∞). But Q (1) = (1 + t) (1 − 3t) − 1 ≤ 0. So
that Q has a unique root t 1 in (0, 1] and a unique root t 2 in [1, ∞), because
3
Q (0 + ) = (1 + t) (1 − 3t) > 0 and Q (∞) = ∞ > 0. Also, because 1+t ≥ 1 is root
for Q, we deduce that t 2 = 1 + t. We form the Rolle’s sequence for the function h
thus: 0 + < t 1 ≤ 1+t < ∞. Since h (0 + ) = −∞ < 0, h (∞) = ∞ > 0 and h admits
at least 3 positive roots, by a consequence of Rolle’s Theorem we deduce that
2
h (t 1 ) ≥ 0 and h (1 + t) ≤ 0. By h (1 + t) ≤ 0 we obtain that s ≤ 4(1 + t) (1 − 2t).
Let’s note that if a = b = c = 1 + t and d = 1 − 3t, then a + b + c + d = 4,
3 2
abcd = (1 + t) (1 − 3t) and abc + abd + acd + bcd = 4(1 + t) (1 − 2t). So indeed,
2
max (abc + abd + acd + bcd) = 4(1 + t) (1 − 2t).

2
2
By Claim 4, (a + b + c + d) + k abc+abd+acd+bcd ≥ 16 + k(1 + t) (1 − 2t)
a+b+c+d
(16+k)(1+t)(1−3t) (16+k)(a+b+c+d)abcd
and ≥ , with equality for a = b = c = 1 + t and
1−2t abc+abd+acd+bcd
2
d = 1 − 3t, so from (2) we get 16 + k(1 + t) (1 − 2t) ≥ (16+k)(1+t)(1−3t) . We have:
1−2t
2

16 + k(1 + t) (1 − 2t) ≥ (16+k)(1+t)(1−3t) ∀t ∈ 0, 1 ⇔ 48t 2 ≥ −k (1 + t) · 4t 3
1−2t 3 1−2t 1−2t
1 48 1 48
∀t ∈ 0, ⇔ ≥ −k ∀t ∈ 0, . Hence −k ≤ inf = 27.
3 4t(1+t) 3 1 4t(1+t)
t∈(0, )
3
In conclusion, if −27 ≤ k < −16 then (2) holds for all a, b, c, d > 0 and hence
1
(1) holds for all a, b, c, d > 0 satisfying a + b + c + d = 1 + 1 + 1 + .
a b c d
4
Let k < −27. We define on (0, ∞) the function
(16+k)(a+b+c+d)abcd
2
ϕ (a, b, c, d) = (a + b + c + d) + k abc+abd+acd+bcd − .
a+b+c+d abc+abd+acd+bcd
Then lim ϕ (1, 1, 1, x) = 9 + k 3 < 0 ⇒ ∃x > 0 for which ϕ (1, 1, 1, x) < 0 ⇒
x&0
ϕ (α, α, α, αx) < 0 ∀α > 0. Let us find α > 0 for which α + α + α + αx = 1 + 1 +
α α
q q q q q
1 + 1 ⇒ α = 3x+1 ⇒ ϕ 3x+1 , 3x+1 , 3x+1 , x · 3x+1 < 0. We
α αx x(x+3) x(x+3) x(x+3) x(x+3) x(x+3)
q q
1
1
set a = b = c = 3x+1 and d = x · 3x+1 . Then a + b + c + d = 1 + + + 1
x(x+3) x(x+3) a b c d
2
and (a + b + c + d) + kabcd < 16 + k.
In conclusion, the required set of admissible k equals the interval [−27, 48].