Page 24 - MATINF Nr. 9-10

24 M. Bencze

sin A R 3R √
P P
Proof. 1) ≤ √ = √ ≤ 3.
2
2
1 + cos B + cos C 2 R − r 2 2 R − r 2
2
2
2
sin A R s
P P
2) ≤ √ sin A = √ .
2
2
1 + cos B + cos C 2 R − r 2 2 R − r 2
2
2
A
sin A sin 2
P 2 R P 2 A 2R − r
3) ≤ √ sin = √ .
2
2
1 + cos B + cos C 2 R − r 2 2 4 R − r 2
2
2
A
sin A cos 2
P 2 R P 2 A 4R + r
4) ≤ √ cos = √ .
1 + cos B + cos C 2 R − r 2 2 4 R − r 2
2
2
2
2
All the next corollaries can be proved in a similar manner as Corollary 1.
Corollary 2. In any acute triangle ABC hold the following inequalities:
…
sin 2A R + r
P
1) ≤ ;
2
2
1 + cos B + cos C R − r
2
2
2
tg A R (s + r − 4R )
P
2) ≤ √ î ó.
2
2
1 + cos B + cos C 2 R − r 2 s − (2R + r) 2
2
2
Corollary 3. In any triangle ABC hold the following inequalities:
2
r b r c sin A s R
P
1) ≤ √ ;
3
2
1 + cos B + cos C 2 R − r 2
2
h b h c sin A s r
P
2) ≤ √ .
2
2
1 + cos B + cos C R − r 2
2
Corollary 4. In any triangle ABC hold the following inequalities:
2 2
a r a 2sR (2R − r)
P
1) ≤ √ ;
2
2
1 + cos B + cos C R − r 2
2
2
a 2 2R (4R + r)
P
2) ≤ √ .
2
2
2
(1 + cos B + cos C)r a s R − r 2
Theorem 2. In any non-obtuse triangle ABC hold the following inequalities
2
sin C
2
2
cos A + cos B ≥ ≥ 2 cos A cos B
1 + cos C
(and their permutations). The second inequality is valid in any triangle.
Proof. We have
2
2
2
2
1 − 2 cos A cos B cos C = cos A + cos B + cos C ≥ 2 cos A cos B + cos C, so
2
sin C ≥ 2 cos A cos B(1 + cos C), and hence
2
sin C
≥ 2 cos A cos B.
1 + cos C
In another way
X 2 X 2 2 2
1 = cos A + 2 cos A cos B cos C ≤ cos A + cos C cos A + cos B , so