Page 29 - MATINF Nr. 8

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is equivalent to
2
4y + 4y − 1
− ≥ 0.
2k
√ √
2
As R ≥ 1 + 2, or 0 ≤ y ≤ 2−1 , we have that 4y + 4y − 1 ≤ 0, and thus our claim follows.
r 2
√
Equality holds if and only if y = 2−1 and x = 1 + y, or if and only if triangle ABC is
2
π
right-isosceles, e.g. A = π and B = C = .
2 4
√
Moving on, it remains to show our inequality under the condition that 2 ≤ R < 1 + 2.
r
This can be done using the equivalent form of Walker’s inequality written in terms of x and y,
namely:
…
1
x ≥ + 4y + 3y .
2
2
In other words, it is enough to prove that

! 2
… Å ã 2
1 y 2 y 1
2
+ 4y + 3y − − (1 + y) − ≥ −
2 k k 2k
√ √
1
holds for all y so that 2−1 < y ≤ , where k ≥ 3 3 − 4.
2 4
After expanding and rearranging, the target inequality reduces to:
√ p
2
k(4y + 4y − 1) + 1 ≥ 2y 2 · 6y + 8y + 1.
2

2
Since 4y + 4y − 1 > 0 in the specified range of y, we can further square both sides and
rearrange to obtain
Ä √ √
î ä ó
2
(1 − 4y) 4y 3 3 − 5 + 15 3 − 26 4y + 4y − 1 ≥ 0,
which is actually the same inequality that we came across in the previous case. This is once
again true as each of the three factors are nonnegative, so we are done. (To be more precise,
note that the last two factors are actually positive.)

Equality holds if and only if 1 − 4y = 0, or, equivalently, if and only if triangle ABC is
equilateral.

5 Final remarks

Coming back to the original problem posed at the beginning of the article, we see that

 Ä √ ä
 cos 3 π + α , if 0 < α ≤ arccot 3 3
3

sup [cos (A + α) cos (B + α) cos (C + α)] =
Ä √ ä
 2 π
sin α cos α, if
 arccot 3 3 < α ≤
3
over all triangles ABC (including non-obtuse-angled triangles).