Page 28 - RMGO 3

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28 Leonard GIUGIUC and Bogdan SUCEAVA

Proof. Consider the polynomial P (x) = (x − a) (x − b) (x − c) (x − d) on (0, ∞).
3
3
2
4
We have: P (x) = x − 4x + mx − sx + (1 − t) (1 + 3t) ∀x > 0, where m and s
are positive with s = abc + abd + acd + bcd.
Consider now the function g (x) = P(x)
2 on (0, ∞). Since g has 4 positive roots,
x
0
by Rolle’s Theorem g has at least 3 positive roots. We have:
3
s 2(1 − t) (1 + 3t)
0
g (x) = 2x − 4 + − , ∀x > 0.
x 2 x 3
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) + s, ∀x > 0. We have:
x
3
0
3
4
h (x) 3x − 4x + (1 − t) (1 + 3t)
= , ∀x > 0.
2 x 2
3
3
4
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 1 = 1 − t. We form
the Rolle’s sequence for the function h thus: 0 + < 1 − t ≤ t 2 < ∞. Since
h (0 + ) = −∞ < 0, h (∞) = ∞ > 0 and h admits at least 3 positive roots, by a
consequence of Rolle’s Theorem we deduce that h (1 − t) ≥ 0 and h (t 2 ) ≤ 0. By
2
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
min (abc + abd + acd + bcd) = 4(1 − t) (1 + 2t).

2 abc+abd+acd+bcd 2
By Claim 2, (a + b + c + d) + k ≥ 16 + k(1 − t) (1 + 2t)
a+b+c+d
and (16+k)(1−t)(1+3t) ≥ (16+k)(a+b+c+d)abcd , with equality for a = b = c = 1 − t
1+2t abc+abd+acd+bcd
2
and d = 1 + 3t, so from (2) we get 16 + k(1 − t) (1 + 2t) ≥ (16+k)(1−t)(1+3t) . We
1+2t
2
have: 16+k(1 − t) (1 + 2t) ≥ (16+k)(1−t)(1+3t) ∀t ∈ [0, 1) ⇔ 48t 2 ≥ k (1 − t)· 4t 3
1+2t 1+2t 1+2t
∀t ∈ [0, 1) ⇔ 48 ≥ k ∀t ∈ (0, 1). Hence k ≤ inf 48 = min 48 = 48,
4t(1−t) 4t(1−t) 4t(1−t)
t∈(0,1) t∈(0,1)
1
at t = .
2
In conclusion, if 0 ≤ k ≤ 48 then (2) holds for all a, b, c, d > 0.
Moreover, if k = 48 then we have equality in (2) at (α, α, α, α) or (α, α, α, 5α)
and permutations, α > 0.
Hence if 0 ≤ k ≤ 48, then (1) holds for all positive real numbers a, b, c and d
1
1
1
satisfying a + b + c + d = 1 + + + . Moreover, if k = 48 then we have equality
√ a b c d

2
2
in (1) at (1, 1, 1, 1) or 10, √ , √ , √ 2 and permutations.
10 10 10