OCTOGON MATHEMATICAL MAGAZINE
Vol. 17, No.1, April 2009, pp 230-236
ISSN 1222-5657, ISBN 978-973-88255-5-0,
www.hetfalu.ro/octogon
230
About a trigonometrical inequality
Mihály Bencze21
ABSTRACT. In this paper we present a trigonometrical inequality and after then
we give some applications.
MAIN RESULTS
Theorem 1. If yk ∈ 0, π2 (k = 1, 2, ..., n) , then
n
X
1
n
p
≤s
n
2
1 + tg yk
k=1
1 P
2
yk
1 + tg n
k=1
Proof. The function f (x) = cos x is concave for x ∈ 0, π2 , therefore from
Jensen’s inequality we get:
!
n
n
X
1X
yk or
cos yk ≤ n cos
n
k=1
k=1
n
X
1
n
p
≤s
n
2
1 + tg yk
k=1
1 P
2
1 + tg n
yk
k=1
Corollary 1.1. If xk > 0 (k = 1, 2, ..., n) , then
n
X
n
1
q
≤s
n
P
1 + x2k
k=1
1
1 + tg 2 n
arctgxk
k=1
Proof. In Theorem 1 we take yk = arctgxk (k = 1, 2, ..., n) .
21
Received: 18.06.2008
2000 Mathematics Subject Classification. 26D15, 51M16
Key words and phrases. Inequalities and inequalities in triangle
231
About a trigonometrical inequality
Corollary 1.2. If xk ∈ (0, 1] (k = 1, 2, ..., n) , then
n
X
k=1
q
1
1 + x2k
≤s
1+
π
4
n
n
Q
2
n
xk
k=1
(k = 1, 2, ..., n) and g (x) = ln tgx is
Proof. If xk ∈ (0, 1], then yk ∈ 0,
concave, therefore from Jensen’s inequality we get
!
n
n
Y
X
1
tgyk ≤ tg n
yk
n
k=1
k=1
and from Theorem 1 we get
n
X
n
n
1
p
≤s
≤s
2
n
1 + tg 2 yk
n
n
Q
k=1
1 P
2
yk
1 + tg n
tgyk
1+
k=1
k=1
after then yields yk = arctgxk (k = 1, 2, ..., n) and we are finish the proof.
Corollary 1.3. If x, y > 0 and xy ≤ 1, then
1
1
2
+p
≤√
2
1 + xy
1+x
1 + y2
2
Proof. If y1 + y2 ≤ π2 , then tgy1 tgy2 ≤ tg 2 y1 +y
, therefore from Corollary
2
1.2 we obtain the result.
√
Corollary 1.4. If xk ∈ [0, 1] (k = 1, 2, ..., n), then
n
X
X
1
1
q
√
≤
1 + x1 x2
1 + x2k
cyclic
k=1
Proof. Using the Corollary 1.3 we obtain

√1 2+√1 2 ≤ √ 2

1+x1 x2

1+x1
1+x2


 − − − − − − − − − − −−
√1 2+√1 2 ≤ √ 2

1+x2 x3

1+x2
1+x3


2
1
 √1
√
+ √ 2 ≤ 1+x
2
x
1+xn
After addition we finish the proof.
1+x1
n 1
232
Mihály Bencze
Corollary 1.5. If xk ∈ [0, 1] (k = 1, 2, ..., n), then
n
2
X
1
2n
q
≤s
21−n
2n
Q
1 + x2k
k=1
xk
1+
k=1
Proof. Using iterative the Corollary 1.3 we get
n
2
X
k=1
1
q
=
1 + x2k
1
1
p
+p
1 + x21
1 + x22
≤√
!
1
1
p
+p
1 + x23
1 + x24
+
!
+ ...
2
2
+√
+ ... ≤
1 + x1 x2
1 + x3 x4
2n
4
+ ... ≤ s
≤p
√
21−n
2n
1 + x1 x2 x3 x4
Q
xk
1+
k=1
Corollary 1.6. In all triangle ABC holds
X
cos
A X
1
q
≤
2
1 + tg A tg B
2
2
Proof. In Corollary 1.4 we take n = 3, x1 = tg A2 , x2 = tg B2 , x3 = tg C2 .
Corollary 1.7. If xk ∈ 0, ln 1 +
n
X
√ 2 (k = 1, 2, ..., n), then
X
1
1
√
≤
chxk
1 + shx1 shx2
k=1
cyclic
Proof. In Corollary 1.4 we take xk → shxk (k = 1, 2, ..., n) .
Theorem 2. If x, y, z > 0 and xyz ≤ 1, then
max
(
2
2
1
2
1
1
√
;√
;√
+p
+√
+√
2
2
1 + xy
1 + yz
1 + zx
1+z
1+x
1 + y2
≤q
1+
3
√
3
xyz
2
)
≤
233
About a trigonometrical inequality
√
√
3
z
1
1
2
= √1+z
+ √2z+u
, then
+ √1+z
xyz and f (z) = √1+xy
2
2
3
′
2
f ( z) = (z − u) 1 − u z + r = 0 if and only if z = u but lim f (x) = 1
Proof. If u =
zց0
and f (u) =
√ 2
1+xy
+
√ 3
1+u2
√ 1
1+z 2
≤
> 2, therefore f (z) ≤ f (u) =
3
q
.
2
√
1+( 3 xyz )
√ 3
1+u2
or
Corollary 2.1. In all triangle ABC holds:
max



2
q
1 + tg A2 tg B2

2
2
C
A
B
≤
+ cos ; q
+ cos ; q
+ cos
2
2
2
1 + tg B tg C
1 + tg C tg A
2
≤ p√
3
√
33s
s2 +
2
√
3
2
2
r2
Proof. In Theorem 2 we take x = tg A2 , y = tg B2 , z = tg C2 etc.
Corollary 2.2. If x, y, z > 0 and xyz ≤ 1, then
√
1
3
1
1
+p
≤q
+√
2
2
2
2
√
1+x
1+z
1+y
1 + 3 xyz
(see [1])
Proof. From Theorem 1 and Theorem 2 we have:
√
1
2
3
1
1
1
+p
≤√
≤q
+√
+√
2
2
2
2
2
√
1
+
xy
1+x
1+z
1+z
1+y
1 + 3 xyz
Corollary 2.3. If x, y, z > 0 and xyz ≤ 1, then
√
1
1
1
3
+√
+p
≤p
1 + x6
1 + z6
1 + y6
1 + x2 y 2 z 2
Proof. In Corollary 2.2 we take x → x3 , y → y 3 , z → z 3 .
Corollary 2.4. In all triangle
ABC holds
√
√
3
P
P
2
33s
A
2R
3
1
√
√√ √
√
2).
cos
≤
≤
1).
√
3
3 2 √
2
3 sr+ 3 2R2
1+sin2 A
s + r 2√
√
3
3
P
P
2
3 16R
3 16R2
q 1
q 1
4).
3).
≤ √√
≤ √√
√
√
3
3
3
3
A
4 A
1+sin
2
r2 +
16R2
1+cos4
2
s2 +
16R2
234
Mihály Bencze
Proof. In Corollary
2.2 we take
(x, y, z) ∈ (sin A, sin B, sin C) ; tg A2 , tg B2 , tg C2 ; sin2 A2 , sin2 B2 , sin2
cos2 A2 , cos2 B2 , cos2 C2 .
Corollary
2.5. In all triangle ABC holdsP
P
2
2
1
q 1
√
1).
≤ √s6s
2).
≤ √s2 6R
2 +r 2
6
A
r 2 +4R4
3).
P
1+sin A
1
q
1+sin12
A
2
≤
√
48R2
4).
r 4 +256R4
P
1+tg 6
C
2
;
2
1
q
1+cos12
A
2
Proof. In Corollary
2.3 we take
(x, y, z) ∈ (sin A, sin B, sin C) ; tg A2 , tg B2 , tg C2 ; sin2
cos2 A2 , cos2 B2 , cos2 C2
≤
2
√ 48R
s4 +256R4
A
2 B
2 C
2 , sin 2 , sin 2
;
Corollary
≤ 1, then
P 1 2.6. If
Px, y,1 z > 0 and xyz
9
√
q
≤
1). 2 √1+xy
+
2
√
1+x2
1+( 3 xyz )
P
P
√ 1
2). 2 √ 1
≤√ 9
+
6
1+x
1+x3 y 3
1+x2 y 2 z 2
Proof. 1). Using the Theorem 2 we get:
2
X
√
X
X
1
1
√
=
+
1 + xy
1 + x2
≤
X
√
2
1
+√
1 + xy
1 + z2
≤
3
9
q
2 = q
2
√
√
1 + 3 xyz
1 + 3 xyz
2). In 1) we take x → x3 , y → y 3 , z → z 3
Corollary 2.7. In all triangle ABC holds√
3
P
P
9 2R2
√ 1
≤ √√
1). 2 √1+sin1A sin B +
√
3
2
3
1+sin A
sr+ 2R2
√
P
P
3s
9
2). 2 q 1 A B + cos A2 ≤ √ √
√
3
3
3). 2
4). 2
P
P
1+tg
2
tg
2
1
q
1+sin2 A
sin2
2
1
q
cos2
1+cos2 A
2
B
2
B
2
+
+
P
P
s2 +
q 1
1+sin4
q 1
1+cos4
A
2
A
2
r2
≤
√
3
16R2
√
3
r 2 + 16R2
√
3
9 16R2
√√
3
3 2 √
s + 16R2
9
≤ √√
3
Proof. In Corollary
2.6 1). we take
(x, y, z) ∈ (sin A, sin B, sin C) ; tg A2 , tg B2 , tg C2 ; sin2
cos2 A2 , cos2 B2 , cos2 C2
Corollary
2.8. In all triangle
ABC holds
P
P
2
1
1
√
√
1). 2
+
≤ √s218R
3
3
6
r 2 +4R4
1+sin A sin B
1+sin A
A
2 B
2 C
2 , sin 2 , sin 2
;
235
About a trigonometrical inequality
2). 2
3). 2
4). 2
P
P
P
1
q
tg 3 B
1+tg 3 A
2
2
1
q
sin6
1+sin6 A
2
1
q
6
cos6
1+cos A
2
+
P
+
B
2
B
2
+
q 1
1+tg 6 A
2
P
P
≤
1
q
1+sin12
1
q
1+cos12
2
√18s
s2 +r 2
A
2
A
2
2
√ 144R
4
r +256R4
≤
≤
2
√ 144R
s4 +256R4
Proof. In Corollary 2.6 1). we take
(x, y, z) ∈ (sin A, sin B, sin C) ; tg A2 , tg B2 , tg C2 ; sin2
cos2 A2 , cos2 B2 , cos2 C2
A
2 B
2 C
2 , sin 2 , sin 2
Theorem 3. If yk ∈ (0, π) (k = 1, 2, ..., n) , then
ntg
n
X
1
n
n
P
yk
tgyk
k=1
p
≤s
n
2
1 + tg yk
k=1
1 P
2
1 + tg n
yk
k=1
Proof. The function f (x) = sin x is concave, therefore from Jensen’s
inequality we have:
n
X
k=1
n
X
n
sin yk ≤ n sin
tgyk
1X
yk
n
k=1
ntg
1
n
!
n
P
or
yk
k=1
p
≤s
n
2y
1
+
tg
k
k=1
1 P
2
yk
1 + tg n
k=1
Corollary 3.1. If yk ∈ 0, π2 (k = 1, 2, ..., n), then
n
1 P
n 1 + tg n
yk
n
X
1 + tgyk
k=1
p
≤ s
n
1 + tg 2 yk
P
k=1
yk
1 + tg 2 n1
k=1
Proof. We adding the inequalities from Theorem 1 and Theorem 3.
Corollary 3.2. If xk > 0 (k = 1, 2, ..., n), then
;
236
Mihály Bencze
n
X
k=1
ntg
xk
1
n
n
P
arctgxk
k=1
q
≤s
n
P
1 + x2k
arctgxk
1 + tg 2 n1
k=1
Proof. In Theorem 3 we take yk = arctgxk (k = 1, 2, ..., n) .
Corollary 3.3. If xk > 0 (k = 1, 2, ..., n), then
n
1 P
arctgxk
n 1 + tg n
n
X
1 + xk
k=1
q
≤ s
n
1 + x2k
k=1
1 P
2
arctgxk
1 + tg n
k=1
Proof. In Corollary 3.1 we take yk = arctgxk (k = 1, 2, ..., n) .
Open Question. If 0 ≤ xk ≤ a (k = 1, 2, ..., n) , then determine all a > 0
and α ∈ R such that
α

1
n 

X
α
1
1


≤
n

α

n
1 + xαk

n 
Q
k=1
xk
1+
k=1
REFERENCES
[1] Arkady Alt, Problem 3329, Crux Mathematicoru 3/2009, pp. 180-181.
[2] Octogon Mathematical Magazine (1993-2009)
Str. Hărmanului 6,
505600 Săcele-Négyfalu
Jud. Braşov, Romania
E-mail: [email protected]