Page 26 - MATINF Nr. 8

26 4 ALTERNATIVE SOLUTION

We will now show that the expression within square brackets is nonpositive. Indeed, since
√
4y ≤ 1 and k ≥ 3 3, we get
Ä 2 ä √ Ä 2 ä
k (4y) − 1 ≤ 3 3 (4y) − 1 ,

so we only need to establish the following inequality:
√
√ Ä 2 ä p
2
3 3 (4y) − 1 + 3 3 ≤ 16y 1 + 2y + 3y .
The latter is equivalent to √
3y 3 ≤ p 1 + 2y + 3y ,
2
which is just
2
−24y + 2y + 1 ≥ 0
after squaring both sides and rearranging. We’ve previously shown that the quadratic expression
factorizes as (1 − 4y)(6y + 1), which is nonnegative, so our inequality holds.

Equality is attained if and only if 4y = 1, or R = 2r, which corresponds to triangle ABC
being equilateral.

î√ ä
3
4.4 inf F = −k in all triangles, k ∈ 3 , ∞
3
3
Note that the inequality F(k) ≥ −k is equivalent to
ñ ô
Å ã 2 Å ã 2
y y
2
2

k k + 1 1 + x − − (1 + y) − ≥ 0.
k k
2
2
2
2
By the other inequality of Gerretsen, i.e. s ≥ 16Rr − 5r , we have that x ≥ 8y − 5y .
Thus, proving that the expression within square brackets is nonnegative boils down to
2 2
Å ã Å ã
y y
p 2
2
1 + 8y − 5y − ≥ (1 + y) + .
k k
Routine calculations allow us to transform the previous inequality into
2y p
2
6y(1 − y) ≥ 8y − 5y ,
k
which holds true if y = 0. Otherwise, we can divide by 2y and square both sides, thus getting
8y − 5y 2
2
k ≥ .
9(y − 1) 2
1
2
Since k ≥ , we are left to prove that
3
1 8y − 5y 2
≥ ,
3 9(y − 1) 2
or

y − 3 y − 1
2 4
≥ 0,
(y − 1) 2
1
which is clearly true given that 0 ≤ y ≤ .
4
Equality holds if and only if triangle ABC is degenerate with two angles equal to 0 (in which
√
case y = 0), or if and only if triangle ABC is equilateral and k = 3 (in which case y = 1/4).
3