OCTOGON MATHEMATICAL MAGAZINE
Vol. 17, No.1, April 2009, pp 70-89
ISSN 1222-5657, ISBN 978-973-88255-5-0,
www.hetfalu.ro/octogon
70
New inequalities for the triangle
Mihály Bencze, Nicuşor Minculete and Ovidiu T. Pop5
ABSTRACT. In this paper we will prove some new inequalities for the triangle.
Among these, we will improve Euler’s Inequality, Mitrinović’s Inequality and
Weitzenböck’s Inequality, thus:
√
1 X p
4
r Fλ (s − a, s − b) (n) ≥ 3 3r;
≥ 2r; s ≥
R≥
P
2
cyclic
Fλ h1a , h1b (n)
cyclic
and
2α
a
+b
2α
2α
+c
α
1 X p
4∆
2α
2α
√
≥
Fλ (a , b ) (n) ≥ 3
,
2
3
cyclic
n
n
where Fλ (x, y) (n) = [(1 + (1 − 2λ) ) x + (1 − (1 − 2λ) ) y] ·
n
n
· [(1 − (1 − 2λ) ) + (1 + (1 − 2λ) ) y] , with λ ∈ [0, 1] , for any x, y ≥ 0 and for all
integers n ≥ 0.
1. INTRODUCTION
Among well known the geometric inequalities, we recall
√ the famous inequality
of Euler, R ≥ 2r, the inquality of Mitrinović, s ≥ 3 3r, and in the year 1919
Weitzenböck published in Mathematische Zeitschrift the following inequality,
√
a2 + b2 + c2 ≥ 4 3∆.
This inequality later, in 1961, was given at the International Mathematical
Olympiad. In 1927, this inequality appeared as the generalization
√ k
2
3 a + bk + c k k
,
∆≤
4
3
in one of the issues of the American Mathematical Monthly. For k = 2, we
obtain the Weitzenböck Inequality.
In this paper we will prove several improvements for these inqualities.
5
Received: 17.03.2009
2000 Mathematics Subject Classification. 26D05, 26D15, 51M04
Key words and phrases. Geometric inequalities, Euler‘s inequality, Mitrinovic‘s
inequality, Weitzenbök inequality.
71
New inequalities for the triangle
2. MAIN RESULTS
In the following, we will use the notations: a, b, c− the lengths of the sides,
ha , hb , hc − the lengths of the altitudes, ra , rb , rc − the radii of the excircles, s
is the semi-perimeter; R is the circumradius, r− the inradius, and ∆− the
area of the triangle ABC.
Lemma 2.1 If x, y ≥ 0 and λ ∈ [0, 1], then the inequality
x+y
2
2
≥ [(1 − λ) x + λy] · [λx + (1 − λ) y] ≥ xy
holds.
Proof. The inequality
x+y 2
≥ [(1 − λ) x + λy] · [λx + (1 − λ) y]
2
is equivalent to
(1 − 2λ)2 x2 − 2 (1 − 2λ)2 xy + (1 − 2λ)2 y 2 ≥ 0,
which means that
(1 − 2λ)2 (x − y)2 ≥ 0,
which is true. The equality holds if and only if λ =
1
2
or x = y.
The inequality
[(1 − λ) x + λy] · [λx + (1 − λ) y] ≥ xy
becomes
λ (1 − λ) x2 − 2λ (1 − λ) xy + λ (1 − λ) y 2 ≥ 0
Therefore, we obtain
λ (1 − λ) (x − y)2 ≥ 0,
which is true, because λ ∈ [0, 1]. The equality holds if and only if
λ ∈ {0, 1} or x = y.
We consider the expression
Fλ (x, y) (n) = [(1 + (1 − 2λ)n ) x + (1 − (1 − 2λ)n ) y] ·
(2.1)
72
Mihály Bencze, Nicuşor Minculete and Ovidiu T. Pop
· [(1 − (1 − 2λ)n ) x + (1 + (1 − 2λ)n ) y] ,
with λ ∈ [0, 1] , for any x, y ≥ 0, and for all integers n ≥ 0.
Theorem 2.2 There are the following relations:
Fλ ((1 − λ) x + λy, λx + (1 − λ) y) (n) = Fλ (x, y) (n + 1) ;
(2.2)
Fλ (x, y) (n + 1) ≥ Fλ (x, y) (n)
(2.3)
(x + y)2 ≥ Fλ (x, y) (n) ≥ 4xy,
(2.4)
and
for any λ ∈ [0, 1] ,for any x, y ≥ 0 and all integers n ≥ 0.
Proof. We make the following calculation:
Fλ ((1 − λ) x + λy, λx + (1 − λ) y) (n) =
= [(1 + (1 − 2λ)n ) ((1 − λ) x + λy) + (1 − (1 − 2λ)n ) (λx + (1 − λ) y)] ·
· [(1 − (1 − 2λ)n ) ((1 − λ) x + λy) + (1 + (1 − 2λ)n ) (λx + (1 − λ) y)] =
= {[1 − λ + (1 − λ) (1 − 2λ)n + λ − λ (1 − 2λ)n ] x+
+ [λ + λ (1 − 2λ)n + 1 − λ − (1 − λ) (1 − 2λ)n ] y}
· {[1 − λ − (1 − λ) (1 − 2λ)n + λ + λ (1 − 2λ)n ] x+
+ [λ − (1 − 2λ)n + 1 − λ + (1 − λ) (1 − 2λ)n ] y} =
=
h
h
i
1 + (1 − 2λ)n+1 x + 1 − (1 − 2λ)n+1 y
i
1 − (1 − 2λ)n+1 x + 1 + (1 − 2λ)n+1 y =
New inequalities for the triangle
73
= Fλ (x, y) (n + 1) ,
so Fλ ((1 − λ) x + λy, λx + (1 − λ) y) (n) = Fλ (x, y) (n + 1) .
We use the induction on n. For n = 0, we obtain the inequality
Fλ (x, y) (1) ≥ Fλ (x, y) (0) .
Therefore, we deduce the following inequality:
4 [(1 − λ) x + λy] · [λx + (1 − λ) y] ≥ 4xy,
which is true, from Lemma 2.1.
We assume it is true for every integer ≤ n, so
Fλ (x, y) (n + 1) ≥ Fλ (x, y) (n)
We will prove that
Fλ (x, y) (n + 2) ≥ Fλ (x, y) (n + 1) .
(2.5)
Using the substitutions x → (1 − λ) x + λy and y → λx + (1 − λ) y in the
inequality (2.3), we deduce
Fλ ((1 − λ) x + λy, λx + (1 − λ) y) (n + 1) ≥
≥ Fλ ((1 − λ) x + λy, λx + (1 − λ) y) (n) ,
so, from equality (2.2), we have
Fλ (x, y) (n + 2) ≥ Fλ (x, y) (n + 1) .
so we obtain (2.6) .
According to inequality (2.3), we can write the sequence of inequalities
Fλ (x, y) (n) ≥ Fλ (x, y) (n − 1) ≥ ... ≥ Fλ (x, y) (1) ≥ Fλ (x, y) (0) = 4xy.
Therefore, we have
Fλ (x, y) (n) ≥ 4xy, for any λ ∈ [0, 1] , x, y ≥ 0 and for all integers n ≥ 0.
74
Mihály Bencze, Nicuşor Minculete and Ovidiu T. Pop
If λ ∈ (0, 1), then 1 − 2λ ∈ (−1, 1) and passing to limit when n → ∞, we
obtain
lim Fλ (x, y) (n) = (x + y)2 .
n→∞
Since the sequence (Fλ (x, y) (n))n≥0 is increasing, we deduce
(x + y)2 ≥ Fλ (x, y) (n) , for any λ ∈ (0, 1) , x, y ≥ 0 and for all integers n ≥ 0.
From the inequalities above we have that
(x + y)2 ≥ Fλ (x, y) (n) ≥ 4xy, for any λ ∈ (0, 1) , x, y ≥ 0 and for all integers n ≥ 0.
If λ = 0 and λ = 1, then Fλ (x, y) (n) = 4xy, so
(x + y)2 ≥ Fλ (x, y) (n) ≥ 4xy.
It follows that
(x + y)2 ≥ Fλ (x, y) (n) ≥ 4xy, for any λ ∈ [0, 1] , x, y ≥ 0 and for all integers n ≥ 0.
Thus, the proof of Theorem 2.2 is complete.
Remark 1. It is easy to see that there is the sequence of inequalities
(x + y)2 ≥ ... ≥ Fλ (x, y) (n) ≥ Fλ (x, y) (n − 1) ≥ ...
... ≥ Fλ (x, y) (1) ≥ Fλ (x, y) (0) = 4xy.
(2.6)
Corollary 2.3. There are the following inequalities:
x+y ≥
p
√
Fλ (x, y) (n) ≥ 2 xy;
x2 + y 2 ≥
x+y+z ≥
p
Fλ (x2 , y 2 ) (n) ≥ 2xy;
√
1 X p
√
√
Fλ (x, y) (n) ≥ xy + yz + zx;
2
cyclic
(2.7)
(2.8)
(2.9)
75
New inequalities for the triangle
x2 + y 2 + z 2 ≥
1 X p
Fλ (x2 , y 2 ) (n) ≥ xy + yz + zx;
2
(2.10)
cyclic
x2 + y 2 + z 2 + xy + yz + zx ≥
1 X
Fλ (x, y) (n) ≥ 2 (xy + yz + zx) (2.11)
2
cyclic
and
(x + y) (y + z) (z + x) ≥
sY
cyclic
Fλ (x, y) (n) ≥ 8xyz,
(2.12)
for any λ ∈ [0, 1] , for any x, y ≥ 0, and for all integers n ≥ 0.
Proof. From Theorem 2.2, we easily deduce inequality (2.7). Using the
substitutions x → x2 and y → y 2 in inequality
p (2.7), we obtain√inequality
(2.8). Similarly to inequality (2.7), x + y ≥ Fλ (x, y) (n) ≥ 2 xy, we can
write the following inequalities:
y+z ≥
p
p
√
√
Fλ (y, z) (n) ≥ 2 yz and z + x ≥ Fλ (z, x) (n) ≥ 2 zx,
which means, by adding, that
x+y+z ≥
√
1 X p
√
√
Fλ (x, y) (n) ≥ xy + yz + zx.
2
cyclic
It is easy to see that, by making the substitutions x → x2 and y → y 2 in
inequality (2.9), we obtain inequality (2.10). Similar to inequality (2.4),
(x + y)2 ≥ Fλ (x, y) (n) ≥ 4xy, we obtain the following inequalities:
(y + z)2 ≥ Fλ (y, z) (n) ≥ 4yz and (z + x)2 ≥ Fλ (z, x) (n) ≥ 4zx.
By adding them, we have inequality (2.11) and by multiplying them, we
obtain inequality (2.12).
Lemma 2.4 For any triangle ABC, the following inequality,
√
ab +
√
bc +
√
ca ≥
4∆
,
R
(2.13)
holds.
Proof. We apply the arithmetic-geometric mean inequality and we find that
√
ab +
√
bc +
√
√
3
ca ≥ 3 abc.
76
Mihály Bencze, Nicuşor Minculete and Ovidiu T. Pop
It is sufficient to show that
√
4∆
3
.
3 abc ≥
R
Inequality (2.14) is equivalent to
27abc ≥
(2.14)
64∆3
,
R3
so
27 · 4R∆ ≥
64∆3
,
R3
which means that
27R4 ≥ 16∆2 .
(2.15)
√
Using Mitrinović’s Inequality, 3 3R ≥ 2s, and Euler’s Inequality R ≥ 2r, we
deduce, by multiplication, that
√
3 3R2 ≥ 4∆.
It follows (2.15) .
Corollary 2.5. In any triangle ABC, there are the following inequalities:
4
r ≥ 2r;
P
1
1
Fλ ha , hb (n)
R≥
(2.16)
cyclic
s≥
√
1 X p
Fλ (s − a, s − b) (n) ≥ 3 3r
2
(2.17)
1 X p
4∆ α
2α
2α
≥
Fλ (a , b ) (n) ≥ 3 √
,
2
3
(2.18)
cyclic
and
2α
a
+b
2α
+c
2α
cyclic
for any λ ∈ [0, 1] , x, y ≥ 0, n ≥ 0 and α is a real numbers.
Proof. Making the substitutions x = h1a , y = h1b and z = h1c in inequality
(2.9), we obtain
1
1
1 X
1
+ + ≥
ha hb hc
2
cyclic
s
Fλ
1 1
,
ha hb
(n) ≥ √
1
1
1
. (2.19)
+√
+√
ha hb
hb hc
hc ha
77
New inequalities for the triangle
According to the equalities
ha =
2∆
2∆
2∆
, hb =
and hc =
,
a
b
c
we have
√
√
√ 1 √
1
1
1
=
+√
+√
ab + bc + ca .
2∆
ha hb
hb hc
hc ha
From Lemma 2.4, we deduce
√
2
1
1
1
≥ .
+√
+√
R
ha hb
hb hc
hc ha
If we use the identity
1
1
1
1
+
+
=
ha hb hc
r
and inequality from above then inequality (2.18) becomes
s 1 1
2
1 X
1
Fλ
(n) ≥ .
≥
,
r
2
ha hb
R
(2.20)
cyclic
Consequently the inequalities (2.16) follows.
If in inequality (2.9) we take x = s − a, y = s − b and z = s − c, then we
deduce the inequality
s≥
1 X p
Fλ (s − a, s − b) (n) ≥
2
cyclic
p
p
p
≥ (s − a) (s − b) + (s − b) (s − c) + (s − c) (s − a).
P p
P √
But, we know the identity
(s − a, s − b) =
bc sin A2 .
cyclic
cyclic
Using the arithmetic-geometric mean inequality, we obtain
cyclic
Hence,
r
r
A
r
B
C
≥ 3 abc sin sin sin = 3 3 4R∆ ·
=
2
2
2
4R
q
√
√
√
√
3
3
3
2
= 3 ∆r = 3 sr ≥ 3 3 3r3 = 3 3r.
X √
A
bc sin
2
3
(2.21)
78
Mihály Bencze, Nicuşor Minculete and Ovidiu T. Pop
p
p
p
√
(s − a) (s − b) + (s − b) (s − c) + (s − c) (s − a) ≥ 3 3r,
(2.22)
which means, according to inequalities (2.21) and (2.22), that
s≥
√
1 X p
Fλ (s − a, s − b) (n) ≥ 3 3r.
2
cyclic
Making the substitutions x = aα , y = bα , and z = cα in inequality (2.9), we
obtain the following inequality:
a2α + b2α + c2α ≥
1 X p
Fλ (a2α , b2α ) (n) ≥ aα bα + bα cα + cα aα .
2
(2.23)
cyclic
Applying the
arithmetic- geometric mean inequality and Pólya-Szegő’s
√
3 2 2 2
4∆
Inequality, a b c ≥ √
, we deduce
3
q
√
α
4∆ α
3
α
3
α α
α α
α α
2
2
2
2
2
2
a b c
,
≥3 √
a b + b c + c a ≥ (a b c ) = 3
3
so
4∆ α
aα bα + bα cα + cα aα ≥ 3 √
.
(2.24)
3
According to inequalities (2.23) and (2.24), we obtain the inequality
4∆ α
1 X p
Fλ (a2α , b2α ) (n) ≥ 3 √
a2α + b2α + c2α ≥
.
2
3
cyclic
Thus, the statement is true.
Remark 2. a) Inequality (2.16) implies the sequence of inequalities
R≥
P
cyclic
r
Fλ
4
≥
1
1
ha , hb
(0)
≥ ... ≥
4
r ≥
P
1
1
Fλ ha , hb (n − 1)
cyclic
4
r ≥ ... ≥ 2r
P
1
1
Fλ ha , hb (n)
cyclic
b) For α = 1 in inequality (2.18), we obtain
(2.25)
79
New inequalities for the triangle
a2 + b2 + c2 ≥
√
1 X p
Fλ (a2 , b2 ) (n) ≥ 4 3∆,
2
(2.26)
cyclic
which proves Weitzenböck’s Inequality, namely
√
a2 + b2 + c2 ≥ 4 3∆.
Corollary 2.6. For any triangle ABC, there are the following inequalities:
1 X p
Fλ (a2 , b2 ) (n) ≥ s2 + r2 + 4Rr,
2 s2 − r2 − 4Rr ≥
2
(2.27)
cyclic
1 X
s − 2r − 8Rr ≥
2
2
2
cyclic
s2 + r2 + 4Rr
4R2
2
r
2
Fλ (s − a) , (s − b)
− 8s2 Rr
≥
2
(n) ≥ r (4R + r) , (2.28)
q
1 X
2s2 r
,
Fλ h2a , h2b (n) ≥
2
R
q
1 X
(4R + r) − 2s ≥
Fλ ra2 , rb2 (n) ≥ s2 ,
2
2
(2.29)
cyclic
2
(2.30)
cyclic
1 X
8R2 + r2 − s2
≥
8R2
2
cyclic
s
Fλ
B
A
, sin4
sin
2
2
4
(n) ≥
s2 + r2 − 8Rr
16R2
(2.31)
and
(4R + r)2 − s2
1 X
≥
4R2
2
cyclic
s
A
B
4
4
Fλ cos
, cos
(n) ≥
2
2
s2 + (4R + r)2
.
8R2
Proof. According to Corollary 2.3 we have the inequality
≥
x2 + y 2 + z 2 ≥
Using the substitutions
1 X p
Fλ (x2 , y 2 ) (n) ≥ xy + yz + zx.
2
cyclic
(2.32)
80
Mihály Bencze, Nicuşor Minculete and Ovidiu T. Pop
(x, y, z) ∈ {(a, b, c) , (s − a, s − b, s − c) , (ha , hb , hc ) , (ra , rb , rc ) ,
A
B
C
B
C
A
sin2 , sin2 , sin2
, cos2 , cos2 , cos2
,
2
2
2
2
2
2
we deduce the inequalities required.
Corollary 2.7. In any triangle ABC, there are the following inequalities:
3s2 − r2 − 4Rr ≥
s2 − r2 − 4Rr ≥
s2 + r2 + 4Rr
4R2
1 X
Fλ (a, b) (n) ≥ 2 s2 + r2 + 4Rr ,
2
(2.33)
cyclic
1 X
Fλ (s − a, s − b) (n) ≥ 2r (4R + r) ,
2
(2.34)
1 X
4s2 r
Fλ (ha , hb ) (n) ≥
2
R
(2.35)
cyclic
2
− 8s2 Rr
≥
cyclic
and
(4R + r)2 − s2 ≥
1 X
Fλ (ra , rb ) (n) ≥ 2s2
2
(2.36)
cyclic
Proof. According to Corollary 2.3, we have the inequality
x2 + y 2 + z 2 + xy + yz + zx ≥
1 X
Fλ (x, y) (n) ≥ 2 (xy + yz + zx) .
2
cyclic
Using the substitutions
(x, y, z) ∈ {(a, b, c) , (s − a, s − b, s − c) , (ha , hb , hc ) , (ra , rb , rc )}
we deduce the inequalities from the statement.
Corollary 2.8. For any triangle ABC there are the following inequalities:
Y p
2s s2 + r2 + 2Rr ≥
Fλ (a, b) (n) ≥ 32sRr,
(2.37)
cuclic
Y p
4sRr ≥
Fλ ((s − a) , (s − b)) (n) ≥ 8sr2 ,
cyclic
(2.38)
81
New inequalities for the triangle
Y p
s2 r s2 + r2 + 4Rr
16s2 r2
F
(h
,
h
)
(n)
≥
≥
,
a
λ
b
R2
R
(2.39)
cyclic
Y p
Fλ (ra , rb ) (n) ≥ 8s2 r,
4s2 R ≥
(2.40)
cyclic
s Y
(2R − r) s2 + r2 − 8Rr − 2Rr2
2 B
2 A
Fλ sin
≥
, sin
(n) ≥
32R3
2
2
cyclic
≥
r2
2R2
(2.41)
and
Y
(4R + r)3 + s2 (2R + r)
≥
32R3
cyclic
s
A
B
s2
2
2
Fλ cos
, cos
(n) ≥
2
2
2R2
(2.42)
Proof. According to Corollary 2.3, we have the inequality
sY
Fλ (x, y) (n) ≥ 8xyz.
(x + y) (y + z) (z + x) ≥
cyclic
Using the substitutions
(x, y, z) ∈ {(a, b, c) , (s − a, s − b, s − c) , (ha , hb , hc ) , (ra , rb , rc ) ,
2 B
2 C
2 B
2 C
2 A
2 A
, sin
, sin
, cos
, cos
, cos
,
sin
2
2
2
2
2
2
we deduce the inequalities required.
Remark 3. From Corollary 2.7, we obtain the inequality
2s s2 + r2 + 2Rr
≥
Y p
Fλ (a, b) (n) ≥ 32sRr ≥
cyclic
≥ 8
Y p
cyclic
Fλ ((s − a) , (s − b)) (n) ≥ 64sr2 .
82
Mihály Bencze, Nicuşor Minculete and Ovidiu T. Pop
We consider the expression
G (x, y) (n) =
xy xn−1 + y n−1
xn+1 + y n+1
(xn + y n )2
,
(2.43)
where x, y > 0 and for all integers n ≥ 0.
Theorem 2.9. For any x, y > 0 and for all integers n ≥ 0, there are the
following relations:
x+y
2
2
≥ G (x, y) (n) ≥ xy
(2.44)
b) G (x, y) (n + 1) ≤ G (x, y) (n) .
(2.45)
a)
and
n
Proof. We take λ = xnx+yn ,for all integers n ≥ 0, in inequality (2.1), because
λ ∈ (0, 1), and we deduce
x + y 2 xy xn−1 + y n−1 xn+1 + y n+1
≥
≥ xy,
2
(xn + y n )2
so,
x+y
2
2
≥ G (x, y) (n) ≥ xy.
To prove inequality (2.45), we can write
G (x, y) (n + 1)
−1=
G (x, y) (n)
=−
(xy)n−1 (x − y)2 x2 + xy + y 2
x2n + x2n−1 y + x2n y 2 + ... + y 2n
(xn+1 + y n+1 )3 (xn−1 + y n−1 )
Consequently, we have
G (x, y) (n + 1) ≤ G (x, y) (n) .
Remark 4. It is easy to see that there is the sequence of inequalities
(x + y)2
= G (x, y) (0) ≥ G (x, y) (1) ≥ ...
4
≤ 0.
New inequalities for the triangle
≥ G (x, y) (n − 1) ≥ G (x, y) (n) ≥ ... ≥ xy.
83
(2.46)
Corollary 2.10
There are the following inequalities:
x+y p
√
≥ G (x, y) (n) ≥ xy;
2
x2 + y 2 p
√
≥ G (x2 , y 2 ) (n) ≥ xy;
2
X p
√
√
√
x+y+z ≥
G (x, y) (n) ≥ xy + yz + zx;
(2.47)
(2.48)
(2.49)
cyclic
x2 + y 2 + z 2 ≥
X p
G (x2 , y 2 ) (n) ≥ xy + yz + zx;
(2.50)
cyclic
X
1 2
G (x, y) (n) ≥ xy + yz + zx (2.51)
x + y 2 + z 2 + xy + yz + zx ≥
2
cyclic
and
1
(x + y) (y + z) (z + x) ≥
8
sY
cyclic
G (x, y) (n) ≥ xyz,
(2.52)
for any x, y > 0 and for all integers n ≥ 0.
Proof. From Theorem 2.9, we easily deduce inequality (2.47). Using the
substitutions x → x2 and y → y 2 in inequality
inequality
p(2.47), we obtain
√
(2.48). Similarly to inequality (2.47), x+y
≥
G
(x,
y)
(n)
≥
xy,
we can
2
write the following inequalities:
√
z+x p
y+z p
√
≥ G (y, z) (n) ≥ yz and
≥ G (z, x) (n) ≥ zx,
2
2
which means, by adding, that
x+y+z ≥
X p
√
√
√
G (x, y) (n) ≥ xy + yz + zx.
cyclic
It is easy to see that by making the substitutions x → x2 and y → y 2 in
inequality (2.49), we obtain inequality (2.50). Similarly to inequality (2.44),
x+y 2
≥ G (x, y) (n) ≥ xy, we obtain the following inequalities:
2
84
Mihály Bencze, Nicuşor Minculete and Ovidiu T. Pop
y+z
2
2
≥ G (y, z) (n) ≥ yz and
z+x
2
2
≥ G (z, x) (n) ≥ zx.
By adding them, we have inequality (2.51) and by multiplying them, we
obtain inequality (2.52).
Corollary 2.11. In any triangle ABC , there are the following inequalities:
2
r ≥ 2r;
P
1
1
G ha , hb (n)
R≥
(2.53)
cyclic
s≥
X p
√
G (s − a, s − b) (n) ≥ 3 3r
(2.54)
X p
4∆ α
2α
2α
,
≥
G (a , b ) (n) ≥ 3 √
3
cyclic
(2.55)
cyclic
and
2α
a
+b
2α
+c
2α
for any n ≥ 0 and for every real numbers α.
Proof. Making the substitutions x = h1a , y = h1b and z =
(2.49), we obtain
1
hc
in inequality
s 1 1
1
1
1
. (2.56)
+√
+√
G
,
(n) ≥ √
ha hb
ha hb
hb hc
hc ha
cyclic
X
1
1
1
+
+
≥
ha hb hc
From inequality (2.13), we have
√
1
2
1
1
≥ ,
+√
+√
R
ha hb
hb hc
hc ha
and from the identity
1
1
1
1
+
=
+
ha hb hc
r
we deduce
X
1
≥
r
cyclic
s 1 1
2
G
,
(n) ≥ .
ha hb
R
(2.57)
New inequalities for the triangle
85
Consequently
R≥
2
r ≥ 2r.
P
1
1
G ha , hb (n)
cyclic
If in inequality (2.49) we take x = s − a, y = s − b and z = s − c, then we
deduce the inequality
s≥
X p
cyclic
G (s − a, s − b) (n) ≥
p
(s − a) (s − b)+
p
p
+ (s − b) (s − c) + (s − c) (s − a).
But
(2.58)
p
p
p
√
(s − a) (s − b) + (s − b) (s − c) + (s − c) (s − a) ≥ 3 3r,
which means that
s≥
X p
cyclic
√
G (s − a, s − b) (n) ≥ 3 3r.
Making the substitutions , and in inequality (2.49), we obtain the following
inequality:
a2α + b2α + c2α ≥
X p
G (a2α , b2α ) (n) ≥ aα bα + bα cα + cα aα .
cyclic
Applying the
arithmetic-geometric mean inequality and Pólya-Szegö’s
√
3 2 2 2
4∆
Inequality, a b c ≥ √
, we deduce
3
4∆ α
.
aα bα + bα cα + cα aα ≥ √
3
Therefore
s≥
X p
cyclic
√
G (s − a, s − b) (n) ≥ 3 3r.
Remark 5. a) Inequality (2.53) implies the sequence of inequalities
R ≥ ...
2
2
r r ≥
≥ ...
P
P
1
1
1
1
G ha , hb (n)
G ha , hb (n − 1)
cyclic
cyclic
(2.59)
86
Mihály Bencze, Nicuşor Minculete and Ovidiu T. Pop
≥
P
cyclic
r
2
1
1
ha , hb
(0)
≥ 2r
(2.60)
b) For α = 1 in inequality (2.55), we obtain
a2 + b2 + c2 ≥
X p
√
G (a2 , b2 ) (n) ≥ 4 3∆,
(2.61)
cyclic
which proves Weitzenböck’s Inequality, namely
√
a2 + b2 + c2 ≥ 4 3∆.
Corollary 2.12. For any triangle ABC , there are the following inequalities:
X p
2 s2 − r2 − 4Rr ≥
G (a2 , b2 ) (n) ≥ s2 + r2 + 4Rr,
(2.62)
cyclic
2
2
s − 2r − 8Rr ≥
X
cyclic
s2 + r2 + 4Rr
4R2
2
r G (s − a)2 , (s − b)2 (n) ≥ r (4R + r) ,
− 8s2 Rr
≥
(4R + r)2 − 2s2 ≥
X
8R2 + r2 − s2
≥
2
8R
cyclic
X q
2s2 r
,
G h2a , h2b (n) ≥
R
(2.63)
(2.64)
cyclic
X q
G ra2 , rb2 (n) ≥ s2
(2.65)
cyclic
s s2 + r2 − 8Rr
4 A
4 B
G sin
, sin
(n) ≥
2
2
16R2
(2.66)
and
X
4 (R + r)2 − s2
≥
4R2
cyclic
s A
B
s2 + (4R + r)2
4
4
G cos
, cos
. (2.67)
(n) ≥
2
2
8R2
Proof. According to Corollary 2.10, we have the inequality
x2 + y 2 + z 2 ≥
X p
G (x2 , y 2 ) (n) ≥ xy + yz + zx.
cyclic
87
New inequalities for the triangle
Using the substitutions
(x, y, z) ∈ {(a, b, c) , (s − a, s − b, s − c) , (ha , hb , hc ) , (ra , rb , rc ) ,
2 A
2 A
2 B
2 C
2 B
2 C
sin
, cos
,
, sin
, sin
, cos
, cos
2
2
2
2
2
2
we deduce the inequalities required.
Corollary 2.13. In any triangle ABC there are the following inequalities:
X
1
G (a, b) (n) ≥ s2 + r2 + 4Rr,
3s2 − r2 − 4Rr ≥
2
(2.68)
cyclic
X
1 2
G (s − a, s − b) (n) ≥ r (4R + r) ,
s − r2 − 4Rr ≥
2
(2.69)
cyclic
s2 + r2 + 4Rr
8R2
2
X
2s2 r
R
(2.70)
i
X
1h
G (ra , rb ) (n) ≥ s2 .
(4R + r)2 − s2 ≥
2
(2.71)
≥
cyclic
G (ha , hb ) (n) ≥
and
cyclic
Proof. According to Corollary 2.10, we have the inequality
X
1 2
G (x, y) (n) ≥ xy + yz + zx.
x + y 2 + z 2 + xy + yz + zx ≥
2
cyclic
Using the substitutions
(x, y, z) ∈ {(a, b, c) , (s − a, s − b, s − c) , (ha , hb , hc ) , (ra , rb , rc )} ,
we deduce the inequalities from the statement.
Corollary 2.14. For any triangle ABC there are the following inequalities:
Y p
1
G (a, b) (n) ≥ 4sRr,
s s2 + r2 + 2Rr ≥
4
(2.72)
cyclic
Y p
1
G ((s − a) , (s − b)) (n) ≥ sr2 ,
sRr ≥
2
cyclic
(2.73)
88
Mihály Bencze, Nicuşor Minculete and Ovidiu T. Pop
Y p
s2 r s2 + r2 + 4Rr
2s2 r2
≥
,
G (ha , hb ) (n) ≥
2
8R
R
(2.74)
cyclic
Y p
1 2
G (ra , rb ) (n) ≥ s2 r,
s R≥
2
(2.75)
cyclic
(2R − r) s2 + r2 − 8Rr − 2Rr2
≥
256R3
s Y
r2
2 B
2 A
G sin
(n) ≥
, sin
≥
2
2
16R2
(2.76)
cyclic
and
Y
(4R + r)3 + s2 (2R + r)
≥
3
256R
cyclic
s s2
B
A
2
2
, cos
(n) ≥
.
G cos
2
2
16R2
(2.77)
Proof. According to Corollary 2.10, we have the inequality
sY
1
G (x, y) (n) ≥ xyz.
(x + y) (y + z) (z + x) ≥
8
cyclic
Using the substitutions
(x, y, z) ∈ {(a, b, c) , (s − a, s − b, s − c) , (ha , hb , hc ) , (ra , rb , rc ) ,
2 B
2 C
2 A
2 B
2 C
2 A
, sin
, sin
, cos
, cos
, cos
,
sin
2
2
2
2
2
2
we deduce the inequalities required.
Remark 6. From Corollary 2.13, we obtain the inequality
Y p
1
G (a, b) (n) ≥ 4sRr ≥
s s2 + r2 + 2Rr ≥≥
4
cyclic
≥8
Y p
G ((s − a) , (s − b)) (n) ≥ 8sr2 .
cyclic
(2.78)
New inequalities for the triangle
89
REFERENCES
[1] Bencze M. and Minculete N., New refinements of some geometrical
inequalities, Octogon Mathematical Magazine, Vol. 16, no.2, 2008.
[2] Botema, O., Djordjević R. Z., Janić, R. R., Mitrinović, D. S. and Vasić, P.
M., Geometric Inequalities, Gröningen,1969.
[3] Mitrinović, D. S., Analytic Inequalities, Springer Verlag Berlin,
Heidelberg, New York, 1970.
[4] Minculete, N. and Bencze, M., A Generalization of Weitzenböck’s
Inequality, Octogon Mathematical Magazine Vol. 16, no.2, 2008.
[5] Minculete, N., Teoreme şi probleme specifice de geometrie, Editura
Eurocarpatica, Sfântu Gheorghe, 2007 (in Romanian).
National College ”Aprily Lajos”
3 După Ziduri Street
500026 Braşov, Romania
E-mail: benczemihaly@yahoo.com
Dimitrie Cantemir University,
107 Bisericii Române Street
500068 Braşov, Romania
E-mail: minculeten@yahoo.com
National College ”Mihai Eminescu”
5 Mihai Eminescu Street,
440014 Satu Mare, Romania
E-mail: ovidiutiberiu@yahoo.com