Page 19 - MATINF Nr. 8

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3 Solution

We will begin by introducing some notation that will be useful in determining both the upper
and lower bounds.

3.1 Preliminaries

By denoting x = cot A, y = cot B, z = cot C, we notice that xy + yz + zx = 1. Moreover, x + y,
y + z, z + x are positive, and as a result x + y + z > 0. This further implies that
√
x + y + z ≥ 3

2
as (x + y + z) ≥ 3(xy + yz + zx) is true for all real numbers x, y, z.
In addition, we can express the cosines and sines of all angles of triangle ABC as follows:
x x 1 1
cos A = √ = p ; sin A = √ = p ;
2
2
x + 1 (x + y)(x + z) x + 1 (x + y)(x + z)
y y 1 1
; ;
cos B = p = p sin B = p = p
2
2
y + 1 (y + z)(y + x) y + 1 (y + z)(y + x)
z z 1 1
cos C = √ = p ; sin C = √ = p .
2
2
z + 1 (z + x)(z + y) z + 1 (z + x)(z + y)
This allows us to rewrite the target product as:
(kx − 1)(ky − 1)(kz − 1)
F(k) = .
(x + y)(y + z)(z + x)

3.2 Upper bounds

Let p = x + y + z be fixed and let r = xyz. We observe that

2
3
(kx − 1)(ky − 1)(kz − 1) k r − k + kp − 1
= := f(r),
(x + y)(y + z)(z + x) p − r
where the domain of the function f is a (possibly degenerate) interval.
Taking the first derivative we see that

2
(kp − 1)(k + 1)
′
f (r) = ,
(k − r) 2
√ √
which is nonnegative as k ≥ 3 and p ≥ 3 imply that kp − 1 ≥ 0. In other words, f is an
3
increasing function of r.
Ä √ ó
2
Now, consider p = 3t +1 , where t is fixed in 0, 3 . It is well known (or see Theorem 2 from
2t 3
[1]) that regardless of whether the triangle is non-obtuse-angled or not, the maximum value of r
is attained when (x, y, z) is any permutation of
Å 2 ã
1 − t
t, t, .
2t