Page 26 - MATINF Nr. 9-10

26 M. Bencze

Corollary 8. In any acute triangle ABC hold the following inequalities:

sin 2A s
P
1) ≤ p ;
2
2
1 + sin B + sin C 3r(2R − r)
A
cos A sin 2 …
P 2 1 2R − r
2) ≤ ;
2
2
1 + sin B + sin C 4 3r
A
cos A cos 2
P 2 4R + r
3) ≤ p ;
2
2
1 + sin B + sin C 4 3r(2R − r)
2
r b r c cos A s R
P
4) ≤ p .
2
2
1 + sin B + sin C 2 3r(2R − r)
Theorem 5. In any triangle ABC holds the following inequality:
A B C √
 
cos cos cos
 
  2R
max 2 , 2 , 2 ≤ √ .
 2 B 2 C 2 B 2 A 2 A 2 B  2 2R − r
 1 + sin + sin 1 + sin + sin 1 + sin + sin 
2 2 2 2 2 2
A 2R − r
P 2
Proof. We have sin = . It follows that
2 2R
A A B C 2R − r A
cos 2 + sin 2 + sin 2 + sin 2 = + cos 2 , so
2 2 2 2 2R 2
…
B C 2R − r A 2R − r A
1 + sin 2 + sin 2 = + cos 2 ≥ 2 · cos , and hence
2 2 2R 2 2R 2
A √
cos 2R
2 ≤ √ .
B C 2 2R − r
1 + sin 2 + sin 2
2 2

Corollary 9. In any triangle ABC hold the following inequalities:
A √
cos
P 2 3 2R
1) B C ≤ √ ;
1 + sin 2 + sin 2 2 2R − r
2 2
A
cos 3
P 2 4R + r
2) B C ≤ p ;
1 + sin 2 + sin 2 2 2R(2R − r)
2 2
A √
sin
P 2 (4R + r) 2R
3) B C ≤ √ ;
1 + sin 2 + sin 2 2s 2R − r
2 2
A √
cos 2
P 2 s 2R
4) Å B C ã A ≤ √ .
1 + sin 2 + sin 2 sin 2r 2R − r
2 2 2