Page 27 - MATINF Nr. 9-10

New inequalities in triangle 27

Theorem 6. In any triangle ABC holds the following inequality:

 
A B C √
sin sin sin
 
  2R
max 2 , 2 , 2 ≤ √ .
B C C A A
1 + cos + cos 1 + cos + cos 1 + cos + cos
 2 2 2 2 2 2 B  2 4R + r
 
2 2 2 2 2 2
A 4R + r
P 2
Proof. We have cos = . It follows that
2 2R
A A B C 4R + r A
sin 2 + cos 2 + cos 2 + cos 2 = + sin 2 , so
2 2 2 2 2R 2
…
B C 4R + r A 4R + r A
1 + cos 2 + cos 2 = + sin 2 ≥ 2 · sin , and hence
2 2 2R 2 2R 2
A √
sin
2 ≤ √ 2R .
B C 2 4R + r
1 + cos 2 + cos 2
2 2

Corollary 10. In any triangle ABC hold the following inequalities:
A √
sin
P 2 3 2R
1) ≤ √ ;
B C 2 4R + r
1 + cos 2 + cos 2
2 2
A
sin 3
P 2 2R − r
2) ;
B C ≤ p
2 2R(4R + r)
1 + cos 2 + cos 2
2 2
A √
cos
P 2 s 2R
3) ≤ √ ;
B C 2r 4R + r
1 + cos 2 + cos 2
2 2
A
sin 2 p
P 2 2R(4R + r)
4) Å ã ≤ .
B C A 2s
1 + cos 2 + cos 2 cos
2 2 2
Theorem 7. In any triangle ABC holds the following inequality:
 
A B C √
sin sin sin
 
  2R
max 2 , 2 , 2 ≤ √ .
B C C A A
cos + cos cos + cos cos + cos
 2 2 2 2 2 2 B  2 2R + r
 
2 2 2 2 2 2
Proof. We have
…
B C r A r A 2R + r A
cos 2 + cos 2 = 2 + − cos 2 = 1 + + sin 2 ≥ 2 · sin ,
2 2 2R 2 2R 2 2R 2
A √
sin 2R
and hence 2 ≤ √ .
B C 2 2R + r
cos 2 + cos 2
2 2