Renements of some geometric inequalities
1
Mihály Bencze , Hui-Hua Wu
1
2
and Shan-He Wu
2
Str. Harmanului 6, 505600 Sacele-Négyfalu, Jud. Brasov, Romania
E-mail: [email protected]
2
Department of Mathematics and Computer Science, Longyan University, Longyan Fujian 364012, P.R. China
E-mail: [email protected]
Abstract:
In this paper, based on the improved versions of certain classical analytic inequalities, some
renements of the inequalities involving the semi-perimeter,circumradius and inradius of a triangle are established.
Keywords: geometric inequality; analytic inequality; renement; triangle; semi-perimeter; circumradius;
inradius
2000 AMS Mathematics subject classication:
1
26D05, 26D15, 51M16
Main Results
For a given triangle ABC , we assume that A, B, C denote its angles, a, b, c denote the lengths of
its corresponding sides, s, R and r denote the semi-perimeter, circumradius and inradius of a triangle
respectively. We will customarily use the notations of cyclic sum and cyclic product such as
X
Y
f (a) = f (a) + f (b) + f (c),
f (a) = f (a)f (b)f (b).
The inequalities related to the semi-perimeter,circumradius and inradius of a triangle have been widely
studied in the literature, see for example [113]. The purpose of this paper is to establish some renements of
this type of geometric inequalities. To do this, we will present the improvements of certain classical analytic
inequalities, these inequalities will be used as the main tool in the proof of the renements of geometric
inequalities.
Theorem 1. If x, y, z > 0, then
r
xy + yz + zx
x+y+z
√
(1)
≥
≥ 3 xyz.
3
3
Proof. Using the arithmetic-geometric means inequality gives
x+y+z
3
2
=
≥
=
1 2
x + y 2 + z 2 + 2xy + 2yz + 2zx
9
1
(xy + yz + zx + 2xy + 2yz + 2zx)
9
1
(xy + yz + zx) ,
3
1
we hence have
x+y+z
≥
3
Application 1.
In all triangle
r
xy + yz + zx
≥
3
ABC
r √
3 3 xy · yz · zx
√
= 3 xyz.
3
holds
3
s2 + r2 + 4Rr
2
≥
≥ (4sRr)
3
s 6 r(4R + r) 3
2
≥
≥ sr2
3
3
2
6 2 3 2 2 2
2s r
2s r
s + r2 + 4Rr
≥
≥
6R
3R
R
6 2 3
2
4R + r
s
≥
≥ s2 r
3
3
6 2
3 2R − r
s + r2 − 8Rr
r 4
≥
≥
2
6R
48R
4R
6 2
3 4R + r
s + (4R + r)2
s 4
≥
≥
2
6R
48R
4R
2s
3
6
Proof.
In Theorem 1 we take (x, y, z) ∈ {(a, b, c), (s − a, s − b, s − c), (ha , hb , hc ),
(ra , rb , rc ), (sin2 A2 , sin2 B2 , sin2 C2 ), (cos2 A2 , cos2 B2 , cos2 C2 )} respectively, we have
rP
P
qY
ab
a
≥
≥ 3
a,
3
3
P
P
Q
from 3 a = 2s
ab = s2 + r2 + 4Rr, a = 4sRr, we deduce the inequality (2).
3 ,
rP
P
qY
(s − a)
(s − a)(s − b)
≥
≥ 3
(s − a),
3
3
P
P
Q
from (s−a)
= 3s , (s − a)(s − b) = r(r + 4R), (s − a) = sr2 , we deduce the inequality (3).
3
rP
P
qY
ha
ha hb
≥
≥ 3
ha ,
3
3
P
P
Q
2
2 2
2
2
from
ha = s +r2R+4Rr , ha hb = 2sR r , ha = 2sRr , we deduce the inequality (4).
rP
P
qY
ra
r a rb
≥
≥ 3
ra ,
3
3
P
P
Q
from
ra = 4R + r, ra rb = s2 , ra = s2 r, we deduce the inequality (5).
s
r
P 2C
P 2A
Y
sin 2 sin2 B2
sin 2
A
3
≥
sin2
≥
3
2
3
P
Q
P 2A
2
2
2
−8Rr
r
from
sin 2 = 2R−r
sin2 A2 sin2 B2 = s +r
, sin2 A2 = 16R
2 , we deduce the inequality (6).
2R ,
16R2
2
(2)
(3)
(4)
(5)
(6)
(7)
P
from
P
cos2
A
2
Theorem 2.
=
If
4R+r
2R ,
P
cos2
3
cos2
x, y, z > 0,
s
P
≥
cos2
B
2
cos2
C
2
B
2
cos2
3
≥
r
Y
3
s2 +(4R+r)2 Q
, cos2 A2
16R2
=
=
cos2
A
,
2
s2
16R2 ,
we deduce the inequality (7).
then
(x + y + z)
Proof.
A
2
A
2
1
1 1
+ +
x y z
s
≥3 1+
3
(x + y) (y + z) (z + x)
xyz
!
≥ 9.
(8)
Using the arithmetic-geometric means inequality gives
(x + y + z)
s
3 1+
3
1
1 1
+ +
x y z
y
z
x z
x y
+ + + + +
x x y
y
z
z
y+z
x+z
x+y
+
+
= 3+
x
y
z
r
y
+
z
x
+
z
x
+y
≥ 3+33
·
·
x
y
z
s
!
3 (x + y) (y + z)(z + x)
= 3 1+
,
xyz
=
(x + y) (y + z)(z + x)
xyz
3+

!
≥ 3 1 +
=
s
3

√
√
√
2 xy · 2 yz · 2 zx

xyz
9.
The inequality (8) is proved.
Application 2.
In all triangle
ABC
holds
!
r
2
2
s2 + r2 + 4Rr
3 s + r + 2Rr
≥9
≥3 1+
2Rr
2Rr
!
r
4R + r
3 4R
≥9
≥3 1+
r
r
!
r
2
2
s2 + r2 + 4Rr
3 s + r + 2Rr
≥9
≥3 1+
2Rr
2Rr
!
r
2
2
2
(2R − r) s2 + r2 − 8Rr
3 (2R − r) (s + r − 8Rr) − 2Rr
≥3 1+
≥9
2Rs2
2Rr2
!
r
3
2
(4R + r) s2 + (4R + r)2
3 (4R + r) + s (2R + r)
≥3 1+
≥9
2Rs2
2Rs2
3
(9)
(10)
(11)
(12)
(13)
Proof.
(sin2
In Theorem 2 we take (x, y, z) ∈ {(a, b, c), (s − a, s − b, s − c), (ha , hb , hc ),
(cos2 A2 , cos2 B2 , cos2 C2 )} respectively, we have
sQ
!
X X1
(a + b)
3
Q
≥ 9 ⇐⇒ (9),
(
a)
≥3 1+
c
a
2 B
2 C
A
2 , sin 2 , sin 2 ),
since
P P1
( a)
a =
X
X
(
(s − a))
Q
4Rr+r 2 +s2
,
2Rr
sQ
1
≥3 1+
s−a
3
(a+b)
Q
c
=
2Rr+r 2 +s2
.
2Rr
[(s − a) + (s − b)]
Q
(s − c)
!
≥ 9 ⇐⇒ (10),
since
P
P 1
(s − a) = s,
s−a =
X
X 1
≥ 3(1 +
(
ha )
ha
4R+r
sr ,
sQ
3
Q
[(s−a)+(s−b)]
Q
(s−c)
=
4R
r .
(ha + hb )
Q
) ≥ 9 ⇐⇒ (11),
hc
since
P
X
ha =
s2 +r 2 +4Rr
,
2R
AX 1
sin2
2
sin2
P
1
ha
= 1r ,
sQ
A
2
≥3 1+
Q
(h
Qa +hb )
hc
s2 +r 2 +2Rr
.
2Rr
=
(sin2 A2 + sin2
Q 2C
sin 2
B
2)
!
≥ 9 ⇐⇒ (12),
since
P
sin2 A2
P
X
1
sin2
A
2
=
(2R−r)(s2 +r 2 −8Rr)
,
2Rr 2
AX 1
cos2
2
cos2
Q
(sin2
Q
sQ
A
2
≥ 3(1 +
3
2 B
A
2 +sin 2 )
C
2
sin 2
=
(cos2 A2 + cos2
Q
cos2 A2
(2R−r)(s2 +r 2 −8Rr )−2Rr 2
.
2Rr 2
B
2)
) ≥ 9 ⇐⇒ (13),
since
P
Theorem 3.
cos2
If
A
2
P
1
cos2
x, y, z > 0,
A
2
=
(4R+r)(s2 +(4R+r)2 )
,
2Rs2
(cos2
Q
2
A
2 +cos
cos2 A
2
B
2 )
=
(4R+r)3 +s2 (2R+r)
.
2s2 R
then
2
Proof.
Q
X
x≤
X x2 + y 2
z
≤2
X x3
Using the arithmetic-geometric means inequality gives
4
yz
.
(14)
X x2 + y 2
2xy 2yz
2zx
+
+
z
x
y
xy yz xy zx yz
zx
+
+
+
+
+
z
x
z
y
x
y
r
r
r
xy yz
xy zx
yz zx
2
·
+2
·
+2
·
z
x
z
y
x y
2(y + x + z)
X
2
x.
≥
z
=
≥
=
=
On the other hand, we have
y3
x4 + y 4
x3
+
=
≥
yz
zx
xyz
1
2
2 (x
+ y 2 )2
x2 + y 2
≥
,
xyz
z
thus
2
X x2
yz
≥
X x2 + y 2
z
.
The Theorem 3 is proved.
Application 3.
In all triangle
ABC
holds
a4
.
(15)
c
abc
Putting in the inequality (14) x = a, y = b, z = c, the inequality (15) follows immediately.
4s ≤
Proof.
Theorem 4.
Proof.
If
x, y, z > 0,
X a2 + b2
≤
2
P
then
n√
√ 2 √
√ 2 o
x+y+z √
1
√
√
− 3 xyz ≥ max ( x − y)2 ,
y− z ,
z− x
.
3
3
Direct computation gives
x+y+z √
1 √
√
− 3 xyz − ( x − y)2
3
3
=
≥
=
thus
(16)
√
xy + xy √
− 3 xyz
3
√
√
3
xyz − 3 xyz
0,
z+
√
1 √
x+y+z √
√
− 3 xyz ≥ ( x − y)2 .
3
3
The Theorem 4 is proved.
Application 4.
In all triangle
ABC
holds
n√
√
√
√
√
√ o
2s √
1
3
− 4sRr ≥ max ( a − b)2 , ( b − c)2 , ( c − a)2
3
3
5
(17)
2 √
2 √
√
2
√
√
√
s √
1
3
− sr2 ≥ max
s−a− s−b ,
s−b− s−c ,
s−c− s−a
3
3
r
np
o
2 2
p
p
p
p
p
s2 + r2 + 4Rr
1
3 2s r
−
≥ max ( ha − hb )2 , ( hb − hc )2 , ( hc − ha )2
6R
R
3
n
o
1
4R + r √
3
2
2
2
− s2 r ≥ max (ra − rb ) , (rb − rc ) , (rc − ra )
3
3
(
r
2 2 2 )
2R − r
r2
1
A
B
C
A
B
C
3
−
≥ max
sin − sin
, sin − sin
, sin − sin
6R
16R2
3
2
2
2
2
2
2
4R + r
−
6R
r
3
1
s2
≥ max
16R2
3
(
A
B
cos − cos
2
2
2 2 2 )
B
C
C
A
, cos − cos
, cos − cos
2
2
2
2
(18)
(19)
(20)
(21)
(22)
Proof.
We take (x, y, z) ∈ {(a, b, c), (s − a, s − b, s − c), (ha , hb , hc ), (ra , rb , rc ),
(sin2 A2 , sin2 B2 , sin2 C2 ), (cos2 A2 , cos2 B2 , cos2 C2 )} in Theorem 4 respectively, and apply the following
identities:
P
Q
Q 4sRr,
P a =2s, a =
(s − a) =s, (s − a) = sr2 ,
Q
P
2 2
s2 +r 2 +4Rr
ha = 2sRr ,
2R Q ,
P ha =
r =4R + r, r = s2 r,
P a 2 A 2R−r Qa 2 A
r2
sin 2 = 2R , sin 2 = 16R
2,
P
Q
4R+r
s2
2 A
2 A
cos 2 = 2R , cos 2 = 16R2 ,
we obtain the desired inequalities (17)(22).
Theorem 5.
If
x, y, z > 0,
then
r
max {x, y, z} − min {x, y, z} ≤
Proof.
4p 2
x + y 2 + z 2 − xy − yz − zx
3
(23)
Without loss of generality, we assume that x = max {x, y, z}, z = min {x, y, z}. Then, we have
x2 + y 2 + z 2 − xy − yz − zx =
≥
=
thus
r
x−z ≤
(x − y)2 + (y − z)2 + (z − x)2
2
1
(x
−
y
+
y
−
z)2 + (z − x)2
2
2
3(x − z)2
,
4
4p 2
x + y 2 + z 2 − xy − yz − zx.
3
The inequality (23) is proved.
Application 5.
In all triangle
ABC
holds
r
max {a, b, c} − min {a, b, c} ≤
4p 2
s − 3r2 − 12Rr
3
6
(24)
r
max {s − a, s − b, s − c} − min {s − a, s − b, s − c} ≤
4p 2
s − 3r2 − 12Rr
3
r s
2
4
s2 + r2 + 4Rr
6s2 r
max {ha , hb , hc } − min {ha , hb , hc } ≤
−
3
2R
R
r q
4
2
max {ra , rb , rc } − min {ra , rb , rc } ≤
(4R + r) − 3s2
3
(26)
(27)
r r
4 16R2 + r2 − 3s2 + 8Rr
2 A
2 B
2 C
2 A
2 B
2 C
max sin
, sin
, sin
− min sin
, sin
, sin
≤
2
2
2
2
2
2
3
16R2
r
4
2 A
2 B
2 C
2 A
2 B
2 C
max cos
, cos
, cos
, − min cos
, cos
, cos
≤
2
2
2
2
2
2
3
s
(25)
(28)
2
(4R + r) − 3s2
16R2
(29)
Proof.
We take (x, y, z) ∈ {(a, b, c), (s − a, s − b, s − c), (ha , hb , hc ), (ra , rb , rc ),
(sin2 A2 , sin2 B2 , sin2 C2 ), (cos2 A2 , cos2 B2 , cos2 C2 )} in Theorem 5 respectively, and apply the following
identities:
a2 + b2 + c2 − ab − bc − ca = s2 − 3r2 − 12Rr,
(s − a)2 + (s − b)2 + (s − c)2 − (s − a)(s − b) − (s − b)(s − c) − (s − c)(s − a) = s2 − 3r2 − 12Rr,
2
2 2
2
h2a + h2b + h2c − ha hb − hb hc − hc ha = s +r2R+4Rr − 6sR r ,
2
ra2 + rb2 + rc2 − ra rb − rb rc − rc ra = (4R + r) − 3s2 ,
sin4 A2 + sin4 B2 + sin4 C2 − sin2 A2 sin2 B2 − sin2 B2 sin2
cos4
A
2
+ cos sin4
Theorem 6.
If
B
2
+ cos4
C
2
x, y, z > 0,
− cos2
A
2
cos2
B
2
− cos2
C
2
B
2
− sin2
cos2
C
2
−
16R2 +r 2 −3s2 +8Rr
,
16R2
(4R+r)2 −3s2
2 C
2 A
cos 2 cos 2 =
.
16R2
C
2
sin2
A
2
=
then
3
3
min (x − y)2 , (y − z)2 , (z − x)2 ≤ x2 +y 2 +z 2 −xy−yz−zx ≤ max (x − y)2 , (y − z)2 , (z − x)2 . (30)
2
2
Proof.
Note that
x2 + y 2 + z 2 − xy − yz − zx =
(x − y)2 + (y − z)2 + (z − x)2
,
2
thus we have
3
min (x − y)2 , (y − z)2 , (z − x)2
2
≤ x2 + y 2 + z 2 − xy − yz − zx
≤
3
max (x − y)2 , (y − z)2 , (z − x)2 .
2
The inequality (30) is proved.
Application 6.
In all triangle
ABC
holds
n
o
3
2
2
2
min (a − b) , (b − c) , (c − a)
2
7
n
o
2
2
2
≤ s2 − 3r2 − 12Rr ≤ max (a − b) , (b − c) , (c − a)
3
min (ha − hb )2 ,(hb − hc )2 (hc − ha )2
2
2
2
3
s + r2 + 4Rr
6s2 r
−
≤ max (ha − hb )2 ,(hb − hc )2 (hc − ha )2
≤
2R
R
2
(31)
(32)
3
min (ra − rb )2 ,(rb − rc )2 (rc − ra )2
2
3
2
≤ (4R + r) − 3s2 ≤ max (ra − rb )2 ,(rb − rc )2 (rc − ra )2
(33)
2
(
2 2 2 )
3
2 A
2 B
2 B
2 C
2 C
2 A
min
sin
− sin
, sin
− sin
, sin
− sin
2
2
2
2
2
2
2
(
2 2 2 )
3
16R2 + r2 − 3s2 + 8Rr
2 A
2 B
2 B
2 C
2 C
2 A
≤ max
sin
− sin
, sin
− sin
, sin
− sin
≤
16R2
2
2
2
2
2
2
2
(34)
(
)
2 2 2
3
A
B
B
C
C
A
min
cos2 − cos2
, cos2 − cos2
cos2 − cos2
2
2
2
2
2
2
2
(
2 2 2 )
3
(4R + r)2 − 3s2
2 A
2 B
2 B
2 C
2 C
2 A
≤ max
cos
− cos
, cos
− cos
cos
− cos
(35)
≤
16R2
2
2
2
2
2
2
2
Proof. The proof of inequalities (31)(35) are similar to the proof of inequalities (24)(29).
Theorem 7. If x, y, z > 0, then
X
3 Y
X 3
2
1X
2
x +
xy ≥
x2 + xy + y 2 ≥
xy
3
3
Proof.
(36)
By applying the arithmetic-geometric means inequality, we have
Y
2
x + xy + y
2
P
≤
x2 + xy + y 2
3
!3
X
3
2
1X
2
=
x +
xy .
3
3
p
p
√
Setting a = x2 + xy + y 2 , b = y 2 + yz + z 2 , c = z 2 + zx + x2 , it is easy to verify that there is a
triangle with the sides length a, b, c, according to the above substitution, the right-hand side inequality of
(36) is equivalent to the well-known Pólya-Szegö inequality in a triangle [14]:
2
(abc) ≥
4F
√
3
3
,
where F denotes the area of a triangle whose sides are a, b, c. This completes the proof of Theorem 7.
8
(37)
Application 7.
In all triangle
ABC
holds
Y
1
(5s2 − 3r2 − 12Rr) ≥
a2 + ab + b2 ≥ (s2 + r2 + 4Rr)3
27
(38)
Y
1
(2s2 − 3r2 − 12Rr) ≥
(s − a)2 + (s − a)(s − b) + (s − b)2 ≥ (r2 + 4Rr)3
27
" #3
2
2 6
Y
1
s2 + r2 + 4Rr
6s2 r
2s r
2
−
≥
h2a + ha hb + h2b ≥
27
2R
R
R
(39)
3 Y 2
1 2(4R + r)2 − 3s2 ≥
ra + ra rb + rb2 ≥ s6
27
(41)
(40)
" #3
2
2
3
Y
2R − r
3(s2 + r2 − 8Rr)
s + r2 − 8Rr
1
4 A
2 A
2 B
4 B
2
−
≥
sin
+ sin
sin
+ sin
≥
27
2R
16R2
2
2
2
2
16R2
(42)
" #3
2
2
3
Y
3 s2 + (4R + r)2
1
4R + r
s + (4R + r)2
4 A
2 A
2 B
4 B
2
−
≥
cos
+ cos
cos
+ cos
≥
27
2R
16R2
2
2
2
2
16R2
(43)
Proof. In Theorem 7 we take (x, y, z) ∈ {(a, b, c), (s − a, s − b, s − c), (ha , hb , hc ),
(ra , rb , rc ), (sin2 A2 , sin2 B2 , sin2 C2 ), (cos2 A2 , cos2 B2 , cos2 C2 )} respectively, we have
X 3
1 X 2 X 3 Y 2
2
a +
ab ≥
a + ab + b2 ≥
ab ,
27
from
P
a = 2s,
P
ab = s2 + r2 + 4Rr, we deduce the inequality (38).
i3 Y i3
X
hX
1 h X
2
(s − a)2 +
(s − a)(s − b) ≥
(s − a)2 + (s − a)(s − b) + (s − b)2 ≥
(s − a)(s − b) ,
27
P
P
from
(s − a) = s, (s − a)(s − b) = r(r + 4R), we deduce the inequality (39).
3 Y
3
X
1 X 2 X
2
ha +
ha hb ≥
h2a + ha hb + h2b ≥
ha hb ,
27
P
P
2
s2 +r 2 +4Rr
from
ha =
, ha hb = 2sR r , we deduce the inequality (40).
2R
i3 Y
3
X
1 h X 2 X
2
ra +
ra rb ≥
ra2 + ra rb + rb2 ≥
ra rb ,
27
P
P
from
ra = 4R + r, ra rb = s2 , we deduce the inequality (41).
X
3 Y X
3
X
1
4 A
2 A
2 B
4 A
2 A
2 B
4 B
2 A
2 B
2
sin
+
sin
sin
≥
sin
+ sin
sin
+ sin
≥
sin
sin
,
27
2
2
2
2
2
2
2
2
2
from
P
sin2
A
2
=
2R−r
2R ,
P
sin2
A
2
sin2
B
2
=
s2 +r 2 −8Rr
,
16R2
9
we deduce the inequality (42).
X
3 Y X
3
2
X
1
4 A
2 A
2 B
4 A
2 A
2 B
4 B
2 A
2 B
2
cos
+
cos
cos
≥
cos
+ cos
cos
+ cos
≥
cos
cos
,
27
2
2
2
2
2
2
2
2
2
from
P
cos2
A
2
Theorem 8.
=
If
4R+r
2R ,
P
cos2
x, y, z > 0,
A
2
cos2
B
2
=
s2 +(4R+r)2
,
16R2
we deduce the inequality (43).
then
X xy 2
9xyz
1 X 2 X ≤
≤
x +
xy .
2 (x + y + z)
x+y
4
Proof.
(44)
By applying the arithmetic-geometric means inequality, we have
p
X xy 2
Y 3 xy 2
√
≥3
3
x+y
x+y
3xyz
= Q√
3
x+y
≥
=
1
3
3xyz
P
(x + y)
9xyz
,
2 (x + y + z)
on the other hand, we have
X xy 2
1 X (x + y)2 y
≤
x+y
4
x+y
X
X 1
=
x2 +
xy .
4
This completes the proof of Theorem 8.
Application 8. In all triangle ABC holds
9Rr ≤
X ab2
1
≤ (3s2 − r2 − 4Rr)
a+b
4
X (s − a) (s − b)2
9r2
1
≤
≤ (s2 − r2 − 4Rr)
2
c
4
"
#
2
X ha h2
9s2 r2
1
s2 + r2 + 4Rr
2s2 r
b
≤
≤
−
s2 + r2 + 4Rr
ha + hb
4
2R
R
X ra r 2
9s2 r
1
b
≤
≤
(4R + r)2 − s2
2 (4R + r)
ra + rb
4
"
#
2
X sin2 A sin4 B
1
9r2
2R − r
s2 + r2 − 8Rr
2
2
≤
≤
−
4
2R
16R2
16R(2R − r)
sin2 A2 + sin2 B2
"
#
2
X cos2 A cos4 B
9s2
1
4R + r
s2 + (4R + r)2
2
2
≤
≤
−
16R(4R + r)
4
2R
16R2
cos2 A2 + cos2 B2
10
(45)
(46)
(47)
(48)
(49)
(50)
Proof.
In Theorem 8 we take (x, y, z) ∈ {(a, b, c), (s − a, s − b, s − c), (ha , hb , hc ),
(ra , rb , rc ), (sin2 A2 , sin2 B2 , sin2 C2 ), (cos2 A2 , cos2 B2 , cos2 C2 )} respectively, we have
Q
X ab2
9 a
1 X 2 X
P ≤
≤ (
a +
ab),
2 a
a+b
4
P
P
Q
from
a = 2s, ab = s2 + r2 + 4Rr, a = 4sRr, we deduce the inequality (45).
Q
X (s − a)(s − b)2
X
9 (s − a)
1 X
P
≤
≤ (
(s − a)2 +
(s − a)(s − b)),
2 (s − a)
c
4
P
P
Q
from (s − a) = s, (s − a)(s − b) = r(r + 4R), (s − a) = sr2 , we deduce the inequality (46).
Q
X ha h2
9 ha
1 X 2 X
b
P
≤
≤
ha +
h a hb ,
2 ha
ha + hb
4
P
Q
P
2
2 2
2
2
from
ha = s +r2R+4Rr , ha hb = 2sR r , ha = 2sRr , we deduce the inequality (47).
Q
X ra r 2
9 ra
1 X 2 X
b
P ≤
≤
ra +
ra rb ,
2 ra
r a + rb
4
P
P
Q
2
2
from
ra = 4R + r, ra rb = s , ra = s r, we deduce the inequality (48).
Q
X sin2 A sin4 B
9 sin2 A2
1 X 4A X 2A
2 B
2
2
sin
+
sin
sin
,
≤
P 2A ≤
4
2
2
2
2 sin 2
sin2 A2 + sin2 B2
P 2A
P 2A
Q
2
2
−8Rr
r2
sin 2 sin2 B2 = s +r
, sin2 A2 = 16R
from
sin 2 = 2R−r
2 , we deduce the inequality (49).
2R ,
16R2
Q
X cos2 A cos4 B
X
9 cos2 A2
1 X
4 A
2 A
2 B
2
2
≤
cos
+
cos
cos
,
≤
P
4
2
2
2
cos2 A2 + cos2 B2
2 cos2 A2
2
2 Q
P
P
s2
from
cos2 A2 = 4R+r
cos2 A2 cos2 B2 = s +(4R+r)
, cos2 A2 = 16R
2 , we deduce the inequality (50).
2R ,
16R2
Theorem 9.
Let
E (m) = (
P
xm )
P
1
xm
(m = 1, 2, . . .),
then for
x, y, z > 0
the following inequality
holds
E (m + 1) ≥ E (m) ≥ · · · ≥ E (1) ≥ 9.
Proof.
By using power means inequality and Cauchy-Schwarz inequality [15], we deduce that
E (m + 1)
=
X
xm+1
X
1
xm+1
X
i
m+1
1 m+1
=
(xm ) m
( m) m
x
X
m+1
m+1
m
X
m+1
1
m
1− m+1
m
1−
m
m
≥ 3
x
3
m
x
X X 1 1 X X 1 m1
=
xm
xm
xm
9
xm
X X 1 ≥
xm
xm
= E(m),
hX
11
(51)
and
E(1) =
X X 1 x
≥ 9.
x
The Theorem 9 is proved.
Application 9.
ABC holds
2
s2 − r2 − 4Rr
s2 + r2 + 4Rr − 16s2 Rr
In all triangle
8s2 R2 r2
2
s2 − 2r2 − 8Rr (4R + r) − 2s2
Proof.
≥
s2 + r2 + 4Rr
≥ 9,
2Rr
4R + r
≥
≥ 9.
s2 r2
r
Putting m = 2 in Theorem 9 gives
1
1
1
1
1 1
2
2
2
+ +
≥ 9.
+ 2 + 2 ≥ (x + y + z)
x +y +z
x2
y
z
x y z
(52)
(53)
(54)
In the inequality (54) we take (x, y, z) ∈ {(a, b, c), (s − a, s − b, s − c)}, respectively, it follows that
X X 1 X X 1 a2
≥
a
≥ 9,
(55)
a2
a
X
X
X
X
1
1
(s − a)2
≥
(s
−
a)
≥ 9.
(56)
(s − a)2
(s − a)
In inequalities (55) and (56), we apply the following identities:
2
(s2 − r2 − 4Rr) s2 + r2 + 4Rr − 16s2 Rr
X
X 1
(
a2 )(
)=
,
a2
8s2 R2 r2
X X1
s2 + r2 + 4Rr
(
a)(
)=
,
a
2Rr
2
(s2 − 2r2 − 4Rr) (4R + r) − 2s2
X
X
1
)=
,
(
(s − a)2 )(
(s − a)2
s2 r2
X
X
1
4R + r
(
(s − a))(
)=
,
(s − a)
r
giving the desired inequalities (52) and (53).
Acknowledgments.
University of China.
The research was supported by the Science and Technology Project of Longyan
12
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