Page 29 - RMGO 3

On a Geometric Theorem 29

Shall prove that if k > 48, then (1) does not hold for all positive real numbers
1
a, b, c and d satisfying a + b + c + d = 1 + 1 + 1 + . Indeed, if we put the vector
√ a b c d

2
2
10, √ , √ , √ 2 in (1) we get k ≤ 48, which is false.
10 10 10
Now we prove that if k ∈ [−16, 48] then (1) holds for all positive real numbers
1
a, b, c and d satisfying a + b + c + d = 1 + 1 + 1 + .
a b c d
h i h
2
2
We have: 64 (a + b + c + d) + kabcd − 16 − k − (16 + k) (a + b + c + d) +
i h i
2
48abcd − 64 = (48 − k) (a + b + c + d) − 16abcd .

2
Via Maclaurin’s Inequality, (a + b + c + d) ≥ 16 abc+abd+acd+bcd = 16abcd
a+b+c+d
h i h i
2 2
⇒ 64 (a + b + c + d) +kabcd−16−k ≥ (16 + k) (a + b + c + d) +48abcd−64 .
h
2
2
But 16 + k ≥ 0 and (a + b + c + d) + 48abcd − 64 ≥ 0 ⇒ 64 (a + b + c + d) +
i
2
kabcd − 16 − k ≥ 0 ⇒ (a + b + c + d) + kabcd ≥ 16 + k.
It remains to study the existence of k < −16 for which (1) holds for all positive
1
1
1
real numbers a, b, c and d satisfying a + b + c + d = 1 + + + . As above, shall
a b c d
study first the existence of k < −16 in (2), for all a, b, c, d > 0.
Again, we may assume WLOG in (2) that a + b + c + d = 4.
Via AM-GM we get 0 < abcd ≤ 1.
1 3 1
Claim 3. The function f : 0, → (0, 1] as f (t) = (1 + t) (1 − 3t) ∀t ∈ 0,
3 3
is bijectively decreasing.
2
0

Proof. As f (t) = −12t(1 + t) ≤ 0 ∀t ∈ 0, 1 , we deduce that f is strictly
3
decreasing. On the other hand, f (0) = 1 and f 1 − 0 = 0; f is continuous, hence
3
it is well defined and bijective.
1
Claim 4. Let t ∈ 0, be a fixed real number. We consider the positive numbers
3
3
a, b, c, d such that a + b + c + d = 4 and abcd = (1 + t) (1 − 3t). Then
2
max (abc + abd + acd + bcd) = 4(1 + t) (1 − 2t) .
Proof. Consider the polynomial P (x) = (x − a) (x − b) (x − c) (x − d) on (0, ∞).
3
3
4
2
We have: P (x) = x − 4x + mx − sx + (1 + t) (1 − 3t) ∀x > 0, where m and s
P(x)
are positive with s = abc+abd+acd+bcd. Consider now the function g (x) = x 2
0
on (0, ∞). Since g has 4 positive roots, by Rolle’s Theorem g has at least 3
positive roots. We have:
3
0
g (x) = 2x − 4 + s − 2(1 + t) (1 − 3t) ∀x > 0.
x 2 x 3