Page 20 - MATINF Nr. 8

20 3 SOLUTION

Therefore, the following inequality holds:

(kx − 1)(ky − 1)(kz − 1) Å kt − 1 ã 2
2
≤ k − kt − 2t .
2
(x + y)(y + z)(z + x) t + 1
î √ ó
Consider the function g : 0, 3 → R given by
3
kt − 1
Å ã 2
2
g(t) = k − kt − 2t
2
t + 1
and note that
2
2
2(k + 1)(kt − 1)(1 − 3t )
′
g (t) = .
2
(t + 1) 3
√
Thus, if k ≥ 3, then
® Ç√ å´
3
max g(t) ∈ g(0), g ,
h √ i 3
t∈ 0, 3
3
√ √
and if 3 ≤ k ≤ 3, then
3
max g(t) = g(0)
h √ i
t∈ 0, 3
3
′
as g (t) ≤ 0 in this case.
Consequently,
√
 3 √
 (k− 3)
 , if k ≥ 3 3
8
max g(t) = .
h √ i √
t∈ 0, 3  √
3  3
k, if ≤ k ≤ 3 3
3
√
So, if k ≥ 3 3, then
√ ä
Ä 3
k − 3
sup F = ,
8
which is attained, for example, when triangle ABC is equilateral.
We also proved that
F(k) ≤ k
√ √
holds for all k so that 3 ≤ k ≤ 3 3. Here we can notice that equality is attained for an
3
isosceles triangle whose angle between the two equal sides, say A, approaches to 0 :

lim [(k cos A − sin A) (k cos B − sin B) (k cos C − sin C)] = k · (−1) · (−1) = k.
A→0 +

Therefore, we can conclude that

 √ 3 √
 (k− 3)
 , if k ≥ 3 3
8
sup F = .
√ √

k, if ≤ k ≤ 3 3
 3
3