Page 28 - MATINF Nr. 8

28 4 ALTERNATIVE SOLUTION

Again, after performing some manipulations, we find that the previous inequality is equivalent
to

2
k + 1 Ä √ p √ ä
2
· 16ky + 16ky − 5k − 8y 2 · 6y + 8y + 1 + 3 3 ≥ 0.
2
8
In other words, we need to show that
√ √ p
2

2
k 16y + 16y − 5 + 3 3 ≥ 8y 2 · 6y + 8y + 1
√ √ √
is true under the conditions 2−1 < y ≤ 1 and 3 ≤ k < 3 3 − 4.
2 4 2
2
Now we can see that 16y + 16y − 5 ∈ (−1, 0] , so
Ä √ ä
2
2

k 16y + 16y − 5 ≥ 3 3 − 4 16y + 16y − 5 .
Therefore, we are left to show that
Ä √ ä √ √ p
2

3 3 − 4 16y + 16y − 5 + 3 3 ≥ 8y 2 · 6y + 8y + 1,
2
which, after squaring both sides and rearranging, becomes
Ä √ √
î ä ó
2
(1 − 4y) 4y 3 3 − 5 + 15 3 − 26 4y + 4y − 1 ≥ 0.
√
2
1
As 2−1 < y ≤ , it means that 1 − 4y ≥ 0 and 4y + 4y − 1 > 0. Therefore, it remains to
2 4
prove the following inequality:
Ä √ ä √
4y 3 3 − 5 + 15 3 − 26 ≥ 0.
That is,
√
3 3 − 5
y ≥ ,
8
which is clearly satisfied in this case.
Note that equality holds if and only if triangle ABC is equilateral.

√ ä
2
î
4.6 inf F = − (k−1) in non-obtuse-angled triangles, k ∈ 3 3 − 4, ∞
2
Lastly, we will use similar casework as in the previous subsection to show that F(k) ≥ − (k−1) 2
√ 2
holds provided that k ≥ 3 3 − 4.

Note that the conclusion is tantamount to the following inequality:

y y 1
Å ã 2 Å ã 2
2
x − − (1 + y) − ≥ − .
k k 2k
√
To begin with, we will use x ≥ 1 + y to prove that the above is true whenever R ≥ 1 + 2.
r
Indeed, the inequality
2 2
y y 1
Å ã Å ã
2
1 + y − − (1 + y) − ≥ −
k k 2k