Page 27 - MATINF Nr. 8

√ 3 î √ √
(k− 3) 3 ä
4.5 inf F = in non-obtuse-angled triangles, k ∈ , 3 3 − 4 27
8 3
√ 3 î√
(k− 3) √ ä
4.5 inf F = in non-obtuse-angled triangles, k ∈ 3 , 3 3 − 4
8 3

We will use a similar strategy as in the previous part, except that this time we will make use of
2
2
2
s ≥ 2R + r and Walker’s inequality, i.e. s ≥ 2R + 8Rr + 3r , both of which being true for
non-obtuse-angled triangles. In terms of x and y, the previous inequalities are equivalent to
…
1
2
x ≥ 1 + y and x ≥ + 4y + 3y ,
2
respectively.
Now, we will minimize F(k) via the first inequality, thus getting:

ñ ô
Å ã 2 Å ã 2
y y
2
2

F(k) ≥ k + k k + 1 1 + y − − (1 + y) − .
k k
In what follows we will prove that
√ ä
Ä 3
ñ ô
Å y ã 2 Å ã 2 k − 3
y
2
2
k + k k + 1 1 + y − − (1 + y) − ≥
k k 8
√
holds whenever R ≥ 1 + 2.
r
After a series of manipulations, the previous inequality can be written as
2
k + 1 Ä √ ä
2
· 3 3 − k − 16y − 16y ≥ 0,
8
so it is enough to establish that
√
2
16y + 16y + k ≤ 3 3
√ √ √
is true under the conditions 0 ≤ y ≤ 2−1 and 3 ≤ k < 3 3 − 4.
2 2
2
This follows immediately based on the fact that 16y + 16y attains a maximum of 4 at
√
√ (k− 3) 3 √
y = 2−1 , so we have now proven that F(k) ≥ holds under the condition R ≥ 1 + 2.
2 8 √ r
Notice that there is no equality case as k is strictly less than 3 3 − 4.
»
2
Similarly, we can minimize F(x) via x ≥ 1 + 4y + 3y :
2
 
! 2
… Å ã 2
1 y 2 y
2

2
F(k) ≥ k + k k + 1  + 4y + 3y − − (1 + y) −  .
2 k k
Therefore, we can prove the desired inequality by showing that
  Ä 3
! 2 √ ä
… Å ã 2 k −
1 y y 3
2
2

2
k + k k + 1  + 4y + 3y − − (1 + y) −  ≥
2 k k 8
î √ √ ä h √
holds for all k ∈ 3 , 3 3 − 4 and triangles ABC satisfying R ∈ 2, 1 + 2 .
3 r