18 chuyên đề toán học theo xu hướng hội nhập quốc tế 2013

Mục lục
Lời nói đầu
Nguyen Van Mau
On the solutions of some classes of functional equations with transformed
argument 3
Dam Van Nhi
A new inequality and identity (M,N) 16
Nguyen Dang Phat
Some National Olympiad Problems in plane geometry 26
Le Anh Vinh
Elementary Counting Problems 57
Đặng Huy Ruận
Hai phương pháp giải bài toán trò chơi bốc vật 66
Trần Nam Dũng
Một số phương pháp xây dựng bài toán về dãy số 81
Tạ Duy Phượng
Sơ lược giới thiệu di sản sách toán trong thư tịch Hán Nôm 96
Nguyễn Minh Tuấn
Lời giải cho một lớp các bất đẳng thức đồng bậc 118
Nguyễn Bá Đang
Ứng dụng tính chất tam giác đồng dạng 140
Trần Xuân Đáng
Sử dụng đạo hàm để chứng minh bất đẳng thức 148
Hoàng Minh Quân
Phương trình bậc bốn và hệ thức hình học trong tứ giác hai tâm 156
Nguyễn Thùy Trang
Hàm số mũ và phương trình hàm liên quan 180
Nguyễn Đình Thức
Một số ứng dụng lượng giác trong bài toán dãy số 187
Nguyễn Hoàng Cương
Phép quay véctơ và một số ứng dụng 201

Nguyễn Thị Giang
Tính chia hết trong các bài toán về dãy số
và phương trình hàm trên tập số nguyên 212
Nguyễn Hữu Thiêm
Ánh xạ và một số bài toán liên quan 228
Phạm Văn Nho
Vài nét về lịch sử phát triển lí thuyết số 235
Trần Quang Vinh
Một số phương pháp giải phương trình và hệ phương trình đại số 248
2
ON THE SOLUTIONS OF SOME CLASSES OF FUNCTIONAL
EQUATIONS WITH TRANSFORMED ARGUMENT
NGUYEN VAN MAU
Abstract
We deal with some functional equations with transformed arguments
in real plane. By an algbraic approach we solve some kinds of functional
equations with the reflection arguments.
f(x, y)±f(2p−x, y)±f (x, 2q−y)+f (2p−x, 2q−y) = h(x, y), (x, y) ∈ Ω,
(0.1)
where (p, q) is the symmetric center of the set Ω ⊂ R × R, h(x, y) is
given.
In applications, we formulate the necessary and sufficient condition
for solvability of the following functional equations
f(xy)±f ((1−x)y±f (x(1−y))+f ((1−x)(1−y)) = h(xy), ∀x, y ∈ (0, 1).
and
f(x + y)± f (−x +y) ±f(x− y) + f(−x −y)) = h(x +y), ∀x, y ∈ (−1, 1)
and describe the formulae of the general solution f (xy) and f(x + y), respectively.
1
3
September 25, 2013

1 Representations of some classes of two-variable
functions with reflection argument
In this section we will describe some classes of two-variable functions with
transfromed argument. Namely, we deal with two-variable functions being
skew symmetric about a given point (p, q).
Definition 1.1. Let be given a set Ω := P × Q ⊂ R × R and the point (p, q)
is the center (centeral point) of Ω. Function f(x, y) defined in Ω is said to be
even-even (or even in both variables or skew symmetric) about the point (p, q)
iff
f(2p − x, y) = f (x, y) and f(x, 2q − y) = f(x, y), ∀(x, y) ∈ Ω.
Definition 1.2. Let be given a set Ω := P × Q ⊂ R × R and the point (p, q)
is the center (centeral point) of Ω. Function f(x, y) defined in Ω is said to be
even-odd about the point (p, q) iff
f(2p − x, y) = f (x, y) and f(x, 2q − y) = −f(x, y), ∀(x, y) ∈ Ω.
Remark 1. Similarly, we have the definitions of odd-even and odd-odd fun-
tions.
In a special case, we have
Definition 1.3. Function f(x, y) defined in R × R is said to be even-even iff
f(−x, y) = f (x, y) and f (x, −y) = f (x, y), ∀x, y ∈ R.
The following natural questions arise:
∗
2000 Mathematics Subject Classification. Primary 39B99, 39B62, 39B22, 39B32, 39B52.
4
Problem 1.1. How to describe the two-variable function f(x, y) in the cases
f(x, y) is even-even in both variables (x, y) about the point (p, q), i.e.
f(2p − x, y) = f (x, y) and f(x, 2q − y) = f(x, y), ∀(x, y) ∈ Ω. (1.1)
Solution. Note that
f(2p − x, 2q − y) = f (x, 2q − y) = f (x, y), ∀x, y ∈ Ω,
so we can write
f(x, y) =

1
4
[f(x, y) + f(x, 2q − y) + f(2p − x, y) + f(2p − x, 2q − y)]. (1.2)
Now we prove that the function f(x, y) is even in both variables (x, y) if and
only if there exists a function g(x, y) defined in R × R such that
f(x, y) =
1
4
[g(x, y) + g(x, 2q − y) + g(2p − x, y) + g(2p − x, 2q − y)]. (1.3)
Indeed, if f (x, y) is of the form (1.3) then it is easy to check the conditions
(1.1) are saistified and if f(x, y) is even then it has the form (1.2) and then
the form (1.3) with g = f.
Corollary 1.1. The two-variable function f(x, y) is even in both variables
(x, y), i.e.
f(−x, y) = f (x, y) and f (x, −y) = f (x, y), ∀(x, y) ∈ R. (1.4)
iff it is of the form
f(x, y) =
1
4
[g(x, y) + g(x, −y) + g(−x, y) + g(−x, −y)]. (1.5)
where g(x, y) is an arbitrary function defined in R × R.
Similarly, we have
Problem 1.2. How to describe the two-variable function f(x, y) in the cases
f(x, y) is even-odd in both variables (x, y) about the point (p, q), i.e.
f(2p − x, y) = f (x, y) and f(x, 2q − y) = −f(x, y), ∀(x, y) ∈ Ω. (1.6)
Solution. Note that
f(2p − x, 2q − y) = f (x, 2q − y) = −f (x, y), ∀x, y ∈ Ω,
5
so we can write
f(x, y) =

1
4
[f(x, y) + f(2p − x, y) − f(x, 2q − y) − f(2p − x, 2q − y)]. (1.7)
Now we prove that the function f(x, y) is even-odd in both variables (x, y) if
and only if there exists a function g(x, y) defined in R × R such that
f(x, y) =
1
4
[g(x, y) + g(2p − x, y) − g(x, 2q − y) − g(2p − x, 2q − y)]. (1.8)
Indeed, if f (x, y) is of the form (1.8) then it is easy to check the conditions
(1.6) are saistified and if f (x, y) is even-odd then it has the form (1.7) and
then the form (1.8) with g = f.
Similarly, we can formulate the following representations
Theorem 1.1. The two-variable function f(x, y) in the cases f(x, y) is even-
even in both variables (x, y) about the point (p, q), i.e.
f(2p − x, y) = f (x, y) and f(x, 2q − y) = f(x, y), ∀(x, y) ∈ Ω. (1.9)
is of the form
f(x, y) =
1
4
[g(x, y) + g(2p − x, y) + g(x, 2q − y) + g(2p − x, 2q − y)]. (1.10)
Theorem 1.2. The two-variable function f(x, y) in the cases f(x, y) is even-
odd in both variables (x, y) about the point (p, q), i.e.
f(2p − x, y) = f (x, y) and f(x, 2q − y) = −f(x, y), ∀(x, y) ∈ Ω. (1.11)
is of the form
f(x, y) =
1
4
[g(x, y) + g(2p − x, y) − g(x, 2q − y) − g(2p − x, 2q − y)]. (1.12)
Theorem 1.3. The two-variable function f(x, y) in the cases f(x, y) is odd-

even in both variables (x, y) about the point (p, q), i.e.
f(2p − x, y) = −f (x, y) and f(x, 2q − y) = f(x, y), ∀(x, y) ∈ Ω. (1.13)
is of the form
f(x, y) =
1
4
[g(x, y) − g(2p − x, y) + g(x, 2q − y) − g(2p − x, 2q − y)]. (1.14)
6
Theorem 1.4. The two-variable function f(x, y) in the cases f(x, y) is odd-
ood in both variables (x, y) about the point (p, q), i.e.
f(2p − x, y) = −f (x, y) and f(x, 2q − y) = −f(x, y), ∀(x, y) ∈ Ω. (1.15)
is of the form
f(x, y) =
1
4
[g(x, y) − g(2p − x, y) − g(x, 2q − y) + g(2p − x, 2q − y)]. (1.16)
Now we consider the special case of two-variable function f(x, y) defined in
the set Ω
0
= (0, 1) × (0, 1) in the cases f(x, y) is even - even in both variables
(x, y) about the point

0,
1
2

. So theorem 2.1 can be formulate in the following
form.
Corollary 1.2. The two-variable function f(x, y) in the case when f(x, y) is
even-even in both variables (x, y) about the point


1
2
,
1
2

, i.e.
f(1 − x, y) = f (x, y) and f(x, 1 − y) = f (x, y), ∀x, y ∈ (0, 1) (1.17)
is of the form
f(x, y) =
1
4
[g(x, y) + g(1 − x, y) + g(x, 1 − y) + g(1 − x, 1 − y)]. (1.18)
Corollary 1.3. The two-variable function f(x, y) in the case when f(x, y) is
odd-odd in both variables (x, y) about the point

1
2
,
1
2

, i.e.
f(1 − x, y) = −f (x, y) and f(x, 1 − y) = −f (x, y), ∀x, y ∈ (0, 1) (1.19)
is of the form
f(x, y) =
1
4
[g(x, y) − g(1 − x, y) − g(x, 1 − y) + g(1 − x, 1 − y)]. (1.20)

2 Functional equations for two-variable fun-
tions induced by involutions
In this section we will solve the following functional equations
f(x, y) + f (2p − x, y) + f(x, 2q − y) + f (2p − x, 2q − y) = h(x, y), ∀(x, y) ∈ Ω
(2.1)
7
and
f(x, y) − f(2p − x, y) − f(x, 2q − y) + f(2p − x, 2q − y) = h(x, y), ∀(x, y) ∈ Ω,
(2.2)
where (p, q) is the symmetric center of the set Ω ⊂ R × R, h(x, y) is given.
Denote by X the set of all functions defined on X and X = L
0
(X, X),
where L
0
(X, X) denotes the linear space of all linear operators A : X → X
with dom A = X. It is easy to check that X is an algebra (linear ring) over
field R.
Consider the following linear elements (operators) V and W in X as follows
(V f)(x, y) = f (2p − x, y), (W f)(x, y) = f(x, 2q − y), f ∈ X. (2.3)
It is easy to see that V and W are involution elements, i.e. V
2
= I and
W
2
= I, where I is an identity element of X . Moreover, they are commutative,
i.e. V W = W V.
Now we rewrite (2.1) in the form
Kf := (I + V + W + V W )f = h. (2.4)
Lemma 2.1. Operator K defined by (2.4) is an algebraic element with chara-

teristic polynomial
P
K
(t) = t
2
− 4t. (2.5)
Proof. Note that the following identities hold
V K = K, W K = K, V WK = K.
So K
2
= 4K and K is not a scalar operator, which toghether imply P
K
(t) =
t
2
− 4t, which was to be proved.
Theorem 2.1. The general solution of the homogeneous equation Kf = 0 is
of the form
f(x, y) =
1
4
[3g(x, y)−g(2p−x, y)−g(x, 2q−y)−g(2p−x, 2q−y)], g ∈ X. (2.6)
Proof. By Lemma 2.1, from equality (K − K)f = 0, ∀f ∈ X, we find
1
4

K
2
−
4K


f = 0 ⇔ (K
2
− 4K)f = 0 ⇔ K(K − 4I)f = 0. Hence (K − 4I)X ⊂
ker K. On the other hand, if ϕ ∈ ker K then Kϕ = 0 and K(K − 4I)ϕ =
(K − 4I)Kϕ = 0. It follows ϕ ∈ (4I − K), which was to be proved.
8
Theorem 2.2. The non-homogeneous equation (2.1) (Kf = h) is solvable if
and only if the following condition
Kh = 4h. (2.7)
If it is the case, then the general solution of (2.1) is of the form
f(x, y) = 3g(x, y) − g(2p − x, y) − g(x, 2q − y) − g(2p − x, 2q − y) (2.8)
+
1
4
[3h(x, y) − h(2p − x, y) − h(x, 2q − y) − h(2p − x, 2q − y)], g ∈ X.
Proof. Suppose that the equation (2.1) is solvable and f
0
is a solution. Then
By Lemma 2.1 from equality Kf
0
= h it follows K
2
f
0
= Kh ⇔ 4Kf
0
= Kh ⇔
4h = Kh.
Suppose that the condition (2.7) is satisfied. Write the non-homogeneous

equation (2.1) in the form Kf =
1
4
Kh or in the equivalent form
K

f −
1
4
h

= 0. (2.9)
Theorem 2.1 gives the general solution of (2.9) in the form
f =
1
4
h + (4I − K)ψ ⇔ f =
1
4
h + (4I − K)ψ, ψ ∈ X,
i.e. it has the form (2.8).
Now we consider (2.2). Write it in the form
Lf := (I − V − W + V W )f = h. (2.10)
Lemma 2.2. Operator L defined by (2.4) is an algebraic element with chara-
teristic polynomial
P
L
(t) = t
2
− 4t. (2.11)

Proof. The proof follows from the following identities
−V L = L, −W L = L, V WL = L.
So L
2
= 4L and L is not a scalar operator, which toghether imply P
L
(t) =
t
2
− 4t, which was to be proved.
Theorem 2.3. The general solution of the homogeneous equation Kf = 0 is
of the form
f(x, y) =
1
4
[3g(x, y) + g(2p − x, y) + g(x, 2q − y) − g(2p − x, 2q − y)], g ∈ X.
(2.12)
9
Proof. By the same method as for theorem2.1.
Theorem 2.4. The non-homogeneous equation (2.1) (Lf = h) is solvable if
and only if the following condition
Lh = 4h. (2.13)
If it is the case, then the general solution of (2.1) is of the form
f(x, y) = 3g(x, y) + g(2p − x, y) + g(x, 2q − y) − g(2p − x, 2q − y) (2.14)
+
1
4
[3h(x, y) + h(2p − x, y) + h(x, 2q − y) − h(2p − x, 2q − y)], g ∈ X.
Proof. Suppose that the equation (2.2) is solvable and f
0

is a solution. Then
By Lemma 2.2 from equality Lf
0
= h it follows L
2
f
0
= Lh ⇔ 4Lf
0
= Lh ⇔
4h = Lh.
Suppose that the condition (2.13) is satisfied. Write the non-homogeneous
equation (2.2) in the form Lf =
1
4
Lh or in the equivalent form
L

f −
1
4
h

= 0. (2.15)
Theorem 2.1 gives the general solution of (2.15) in the form
f =
1
4
h + (4I − L)ψ ⇔ f =
1

4
h + (4I − L)ψ, ψ ∈ X,
i.e. it has the form (2.14).
3 Special cases
Now we consider some special cases of equation when q = p =
1
2
. In that case,
the center point of Ω is

1
2
,
1
2

. and (2.1) is of the form
f(x, y)+f (1−x, y)+f (x, 1−y)+f(1−x, 1−y) = h(x, y), ∀x, y ∈ (0, 1). (3.1)
and
f(x, y)−f (1−x, y)−f (x, 1−y)+f (1−x, 1−y) = h(x, y), ∀x, y ∈ (0, 1). (3.2)
In this case, the role of x and y in the left side of (3.1) are the same.
Now return to the function f (t), we can formulate the following
10
www.VNMATH.com
Theorem 3.1. The function f(t) satisfying the conditions
f((1 − x)y) = f(xy), ∀x, y ∈ (0, 1) (3.3)
if and only if there exists a function g(t) such that
f(xy) =
1
4

[g(xy) + g((1 − x)y) + g(x(1 − y)) + g((1 − x)(1 − y))]. (3.4)
Proof. From (3.7) we find f(x(1 − y)) = f(xy) and
f((1 − x)(1 − y)) = f(x(1 − y)) = f (xy), ∀x, y ∈ (0, 1).
So we can write f(t) in the form
f(xy) =
1
4
[f(xy) + f ((1 − x)y) + f(x(1 − y)) + f((1 − x)(1 − y))].
Last equality gives us the proof of the theorem.
Theorem 3.2. The functional equation
f(xy) + f ((1 − x)y) + f(x(1 − y)) + f((1 − x)(1 − y))] = h(xy) (3.5)
is solvable if and only if there exists h(t) satisfying the following condition
h(x(1 − y)) = h(xy), ∀x, y ∈ (0, 1). (3.6)
So now we shall examine the equation (3.6).
Putting x = t, y =
1
2t
, where t ∈

1
2
, 1

into (3.6), we get
h

t −
1
2


= h

1
2

, ∀t ∈

1
2
, 1

.
or
h(x) = h

1
2

, ∀x ∈

0,
1
2

. (3.7)
Similarly, putting xy = t, then x(1 − y) = t
1 − y
y
and
h(t) = h


1 − y
y
t

, ∀y ∈ (0, 1). (3.8)
(3.7) and (3.8) together inply h(t) ≡ const .
So, the necessary condition for solvability of (3.1) if
h(x, y) = h(y, x), ∀x, y ∈ (0, 1). (3.9)
Now we formulate the similar results as in the previous section.
11
Theorem 3.3. The general solution of the homogeneous equation
f(x, y) + f(1 − x, y) + f(x, 1 − y) + f(1 − x, 1 − y) = 0, ∀x, y ∈ (0, 1) (3.10)
is of the form
f(x, y) =
1
4
[3g(x, y)− g(1− x, y) −g(x, 1− y)− g(1− x, 1− y)], g ∈ X. (3.11)
Theorem 3.4. The non-homogeneous equation
f(xy) + f((1 − x)y) + f(x(1 − y)) + f((1 − x)(1 − y)) = h(xy), ∀x, y ∈ (0, 1)
(3.12)
is solvable iff h(t) ≡ const in (0, 1). If it is the case, then the general solution
of (3.14) is of the form
f(xy) =
1
4
[c+3g(xy)−g((1−x)y)−g(x(1−y))−g((1−x)(1−y))], c ∈ R, g ∈ X.
(3.13)
Proof.
Theorem 3.5. The function f(t) is a general solution of the homogeneous

equation
f(xy)+f ((1− x)y)+f(x(1− y))+f ((1− x)(1− y)) = 0, ∀x, y ∈ (0, 1) (3.14)
if and only if there exists a function g(t) such that
f(xy) =
1
4
[3g(xy)−g((1−x)y)−g(x(1−y))−g((1−x)(1−y))], g ∈ X. (3.15)
Remark 2. The homogeneous equation (3.14) was posed by Sahoo and Sander
in 1990 (see [3]-[4]). The continuous solutions were found by Z. Daroczy and
A. Jarai in [5].
Now we deal with the equation induced by addition of arguments. We
consider the special cases of equation when q = p = 0 In that case, the center
point of Ω is (−1, 1) and (2.1) is of the form
f(x, y) + f(−x, y) + f(x, −y) + f(−x, −y) = h(x, y), ∀x, y ∈ (−1, 1). (3.16)
Theorem 3.6. The function f(t) is a general solution of the homogeneous
equation
f(x + y) + f (−x + y) + f (x − y) + f(−x − y)) = 0, ∀x, y ∈ (−1, 1) (3.17)
if and only if f(x) is an odd function in (−2, 2).
12
www.VNMATH.com
Proof. If f is odd in (−2, 2), then f(x − y) = −f(−x + y) and f(−x − y) =
−f(x + y). Hence,
f(x + y) + f (−x + y) + f (x − y) + f(−x − y)) = 0, ∀x, y ∈ (−1, 1).
Conversely, suppose f is a solution of (3.17). Putting x = 0, y = 0 into (3.17)
we find f(0) = 0. Similarly, putting y = x, into (3.15) and ussing the equality
f(0) = 0, we find f(−2x) = −f (2x), i.e. f is odd in (−2, 2).
Theorem 3.7. The non-homogeneous equation
f(x+y)+f (−x+y)+f(x−y)+f (−x−y)) = h(x+y), ∀x, y ∈ (−1, 1) (3.18)
is solvable iff h(t) ≡ const in (−2, 2). If it is the case, then the general solution
of (3.14) is of the form

f(t) =
1
4
h(0) + g(t), (3.19)
where g is an arbitrary odd function in (−2, 2).
Proof. Suppose equation (3.18) is solvable and f is its solution. Putting y = x
and y = −x into (3.18), we find h(2x) = h(0), x ∈ (−1, 1), i.e. h(t) ≡ h(0) in
(−2, 2).
If h(t) ≡ h(0) in (−2, 2) then we can reduce equation (3.18) to the equation
ϕ(x + y) + ϕ(−x + y) + ϕ(x − y) + ϕ(−x − y)) = 0, ∀x, y ∈ (−1, 1), (3.20)
where ϕ(t) = f (x)−
1
4
. Hence, the solution (3.19) follows from theorem 3.6.
Now we return to the equation
f(xy) − f ((1 − x)y) − f(x(1 − y)) + f((1 − x)(1 − y)) = 0. (3.21)
By the same way as previous equations, we have
Theorem 3.8. The function f(t) satisfying the conditions
f((1 − x)y) = −f(xy), ∀x, y ∈ (0, 1) (3.22)
if and only if there exists a function g(t) such that
f(xy) =
1
4
[g(xy) − g((1 − x)y) − g(x(1 − y)) + g((1 − x)(1 − y))]. (3.23)
13
Proof. From (3.23) we find f(x(1 − y)) = −f(xy) and
f((1 − x)(1 − y)) = −f(x(1 − y)) = f (xy), ∀x, y ∈ (0, 1).
So we can write f(t) in the form
f(xy) =
1

4
[f(xy) − f ((1 − x)y) − f(x(1 − y)) + f((1 − x)(1 − y))].
Last equality gives us the proof of the theorem.
Theorem 3.9. The functional equation
f(xy) − f ((1 − x)y) − f(x(1 − y)) + f((1 − x)(1 − y))] = h(xy) (3.24)
is solvable if and only if there exists h(t) ≡ 0.
Theorem 3.10. The non-homogeneous equation
f(xy) − f ((1 − x)y) − f (x(1 − y)) + f((1 − x)(1 − y)) = h(xy), ∀x, y ∈ (0, 1)
(3.25)
is solvable iff h(t) ≡ 0 in (0, 1). If it is the case, then the general solution of
(3.14) is of the form
f(xy) =
1
4
[3g(xy)+g((1−x)y)+g(x(1−y))−g((1−x)(1−y))], g ∈ X. (3.26)
Theorem 3.11. The function f (t) is a general solution of the homogeneous
equation
f(xy)− f ((1−x)y)− f (x(1− y))+f((1− x)(1− y)) = 0, ∀x, y ∈ (0, 1) (3.27)
if and only if there exists a function g(t) such that
f(xy) =
1
4
[3g(xy)+g((1−x)y)+g(x(1−y))−g((1−x)(1−y))], g ∈ X. (3.28)
Remark 3. The equation (3.27) was posed firstly by K. Lajko in [8] for X = R
and then by Sahoo and Sander in 1990 (see [3]-[4]). The differentiable solutions
were found by C.J. Eliezer in [6].
Theorem 3.12. The non-homogeneous equation
f(x+y)−f (−x+y)−f (x−y)+f (−x−y)) = h(x+y), ∀x, y ∈ (−1, 1) (3.29)
is solvable iff h(t) ≡ const in (−2, 2). If it is the case, then the general solution
of (3.27) is of the form

f(t) = c +
1
2
(g(t) − g(−t)), t ∈ (−2, 2), (3.30)
where g is an arbitrary function in (−2, 2), c = f(0).
14
www.VNMATH.com
Proof. Suppose equation (3.31) is solvable and f is its solution. Putting y = x
and y = −x into (3.31), we find h(2x) = −h(0), x ∈ (−1, 1), i.e. h(t) ≡ −h(0)
in (−2, 2). Put x = 0 = y into (3.31) we find h(0) = 0.
So (3.30) is of the form
f(x + y) − f (−x + y) − f (x − y) + f(−x − y)) = 0, ∀x, y ∈ (−1, 1) (3.31)
and it has solution of the form (3.30).
References
[1] T. Acze’l, Lectures on functional equations and their applications, Aca-
demic Press, New York/San Francisco/London, m1966.
[2] M. Kuczma, B. Choczewski, R. Ger, Interative Functional Equa-
tions, Cambridge University Press, Cambridge/New York/Port
Chester/Melbourne/Sydney, 1990.
[3] P.K. Sahoo, T. Riedel, Mean Value Theorems and Functional Equations,
World Scientific, Singapore/New Jersey/London/HongKong, 1998.
[4] B.R. Ebanks, P.K. Sahoo and W. Sander, Determination of measurable
sum from infomation measures satisfying (2, 2)−additivity of degree (α, β),
Radovi Matematicki, vol.6, 77-96, 1990.
[5] Z. Daroczy and A. Jarai, On the measurable solution of a functional equa-
tion of the information theory, Acta Math. Acad Sci. Hungaricae, vol.34,
105-116, 1979.
[6] C.J. Eliezer, A solution to f (1−x)(1−y))+f(xy) = f (x(1−y))+f (1−x)y,
Aequationes Math., vol.46, 301, 1993.
[7] Gy. Maksa, Problem, Aequationes Math., vol.46, 301, 1993.

[8] K. Lajko, What is the general solution to the equation f(1 − x)(1 − y)) +
f(xy) = f(x(1 − y)) + f(1 − x)yAequationes Math., vol.10, 313, 1974.
[9] D. Przeworska - Rolewicz, Algebraic analysis, PWN - Polish Scientific
Publishers and D. Reided Publishing Company, Warszawa - Dordrecht,
1988.
[10] D. Przeworska - Rolewicz and S. Rolewicz, Equations in linear space,
Monografie Matematyezne 47, PWN - Polish Scientific Publishers,
Warszawa, 1968.
[11] Ng. V. Mau, Boundary value problems and controllability of linear systems
with right invertible operators, Dissertationes Math., CCCXVI, Warszawa,
1992.
15
INEQUALITY AND IDENTITY (M, N)
Dam Van Nhi
Pedagogical University Ha Noi
136 Xuan Thuy Road, Hanoi, Vietnam.
Email:
2010 Mathematics subject classification: 26D05,26D15,51M16
Abstract. In this paper we introduce the inequalty and identity (M, N ), of which Hayashi
is a spcial case: Inequality, Identity (M, N ). And after then we will present some inter-
esting applications.
Keywords. Hayashi’s inequality, point, triangle, polygon.
1 Introduction
In Euclidian geometry, the Hayashi’s Inequality in R
2
states: Suppose given a triangle
ABC of the lengths of sides a, b, c. Then, with any point M, we have an inequality
aMB.MC + bMC.MA + cMA.M B  abc
(see [3, pp. 297, 311]). In this paper we propose to give a new inequality and it’s
identity, which is a generalization of the above inequality, and after then we want to

give some interesting applications about triangle.
2 Inequality and Identity (M, N)
Now we prove an inequality, of which Hayashi is a special case. Use this result we can
give some new interesting inequalities. We give the following:
Theorem 2.1. Let A
1
A
2
. . . A
n
be a polygon, s be an integer, s < n, and arbitrary
points N
1
, N
2
, . . . , N
s
, M in euclidean plane R
2
we have the following inequality
s

j=1
MN
j
n

i=1
MA
i


n

k=1
s

j=1
A
k
N
j

i=k
A
k
A
i
.MA
k
, (M, N).
16
www.VNMATH.com
(i) If s = 0, we have Hayashi inequality.
(ii) If n = 3, s = 1, and A, B, C, N belong to the circle with the centere M we have the
inequality aAN + bBN + cCN  4S
ABC
.
Proof. Suppose that A
k
have affixe a

k
, M has affixe z and N
h
affixe z
h
. Using the
Lagrange interpolation formula, we have
s

j=1
(z − z
j
) =
n

k=1
s

j=1
(a
k
− z
j
)

i=k
(a
k
− a
i

)

i=k
(z − a
i
) and
deducing
s

j=1
|z − z
j
|

i=1
|z − a
i
|

n

k=1
s

j=1
|a
k
− z
j
|


i=k
|a
k
− a
i
||z − a
k
|
. From this, we deduce geometric inequal-
ity
s

j=1
MN
j
n

i=1
MA
i

n

k=1
s

j=1
A
k

N
j

i=k
A
k
A
i
.MA
k
.
(i) When s = 0 we have

MN
j
= 1 =

A
k
N
j
and the inequality (M, N) becomes the
Hayashi inequality for the polygon
1
n

i=1
MA
i


n

k=1
1

i=k
A
k
A
i
.MA
k
.
(ii) When n = 3, s = 1 and A, B, C, N belong to the circle with the centere M we have
the inequality
abc
R
 aAN + bBN + cCN or aAN + bBN + cCN  4S
ABC
. .
Remark 2.2. Denote N as the center of circumcircle. From the iequality aAN +bBN +
cCN  4S
ABC
by the inequality (M,N) (ii) we deduce R(a + b + c)  2r(a + b + c) or
R  2r [Euler].
Corollary 2.3. Suppose that O, I and G are respectively the centre of circumcirle and
incircle of ABC. Denote the radii of circumcircles of the triangles GBC, GCA, GAB
by R
1
, R

2
, R
3
, respectivly. Let r
a
, r
b
, r
c
be the the radius of circumcircle of the trian-
gles IBC, ICA, IAB, and let R

1
, R

2
, R

3
be the radii of circumcircles of the triangles
OBC, OCA, OAB, respectivly. We will have
(i) R
2

abc
a + b + c
.
(ii) R
1
+ R

2
+ R
3
 3R (see [1]).
(iii)
r
a
h
a
+
r
b
h
b
+
r
c
h
c

R
r
with h
a
, h
b
, h
c
are the lengths of altitudes of ∆ABC.
(iv)

R

1
x
h
a
+
R

2
y
h
b
+
R

3
z
h
c
 R when ∆ABC is not obtituse and x, y, z are the distances
from O to the 3 sides.
17
Proof. (i) Applying the inequality (M,N) (ii) we obtain aOB.OC+bOC.OA+cOA.OB 
abc or R
2

abc
a + b + c
.

(ii) Applying the inequality (M, N) we obtain aGB.GC + bGC.GA + cGA.GB  abc.
Since aGB.GC = 4R
1
S
GBC
= 4R
1
S
ABC
3
= 4R
1
abc
3.4R
= R
1
abc
3R
, bGC.GA = R
2
abc
3R
and
cGA.GB. = R
2
abc
3R
therefore R
1
abc

3R
+ R
2
abc
3R
+ R
3
abc
3R
 abc or R
1
+ R
2
+ R
3
 3R.
(iii) Applying the inequality (M, N ) we have aIB.IC + bIC.IA + cIA.IB  abc. Since
aIB.IC = 4r
a
S
IBC
= 2r
a
ra = 4
r
a
h
a
rabc
4R

=
r
a
h
a
rabc
R
, bIC.IA =
r
b
h
b
rabc
R
and cIA.IB =
r
c
h
c
rabc
R
there is
r
a
h
a
rabc
R
+
r

a
h
a
rabc
R
+
r
a
h
a
rabc
R
 abc or
r
a
h
a
+
r
b
h
b
+
r
c
h
c

R
r

.
(iv) Applying the inequality (M, N ) we have aOB.OC+bOC.OA+cOA.OB  abc. Since
aOB.OC = 4R

1
S
OBC
= 2R

1
xa = 4R

1
x
h
a
abc
4R
= R

1
x
h
a
abc
R
, bOC.OA = R

2
y

h
b
abc
R
and
cOA.OB = R

3
z
h
c
abc
R
we have
R

1
x
h
a
abc
R
+
R

2
y
h
b
abc

R
+
R

3
z
h
c
abc
R
 abc or
R

1
x
h
a
+
R

2
y
h
b
+
R

3
z
h

c

R.
Proposition 2.4. Suppose given a triangle ABC with the lengths of sides a, b, c respec-
tively and R is the radius of circumcircle of ABC. Let’s I, J
a
, J
b
, J
c
are the centres of
incircle and escribed circles of ∆ABC, respectively. Then, with any point M, we have
(i)
abcMI
MA.MB.MC

aAI
MA
+
bBI
MB
+
cCI
MC
.
(ii)
MI
√
a + b + c
MA.MB.MC


√
b + c − a
√
bcMA
+
√
c + a − b
√
caMB
+
√
a + b − c
√
abMC
.
(iii)
MJ
a
+ M J
b
+ M J
c
MA.MB.MC

AJ
a
+ AJ
b
+ AJ

c
bcMA
+
BJ
a
+ BJ
b
+ BJ
c
caMB
+
CJ
a
+ CJ
b
+ CJ
c
abMC
.
(iv)
MJ
a
.MJ
b
+ M J
b
.MJ
c
+ M J
c

.MJ
a
MA.MB.MC

AJ
a
.AJ
b
+ AJ
b
.AJ
c
+ AJ
c
.AJ
a
bcMA
+
BJ
a
.BJ
b
+ BJ
b
.BJ
c
+ BJ
c
.BJ
a

caMB
+
CJ
a
.CJ
b
+ CJ
b
.CJ
c
+ CJ
c
.CJ
a
abMC
.
Proof. (i) Applying the inequality (M, N ) we have
MI
MA.MB.MC

AI
bcMA
+
BI
caMB
+
CI
abMC
.
(ii) Since IA

2
=
bc(b + c − a)
a + b + c
, IB
2
=
ca(c + a − b)
a + b + c
, IC
2
=
ab(a + b − c)
a + b + c
therefore
MI
√
a + b + c
MA.MB.MC

√
b + c − a
√
bcMA
+
√
c + a − b
√
caMB
+

√
a + b − c
√
abMC
.
18
www.VNMATH.com
(iii) Applying the inequality (M, N) to n = 3, s = 1, we have the three inequalities
MJ
a
MA.MB.MC

AJ
a
bcMA
+
BJ
a
caMB
+
CJ
a
abMC
MJ
a
MA.MB.MC

AJ
b
bcMA

+
BJ
b
caMB
+
CJ
b
abMC
MJ
c
MA.MB.MC

AJ
c
bcMA
+
BJ
c
caMB
+
CJ
c
abMC
.
On adding the three inequalities, we find the inequality
MJ
a
+ M J
b
+ M J

c
MA.MB.MC

AJ
a
+ AJ
b
+ AJ
c
bcMA
+
BJ
a
+ BJ
b
+ BJ
c
caMB
+
CJ
a
+ CJ
b
+ CJ
c
abMC
.
(iii) Applying the inequality (M, N) to n = 3, s = 1, we have the three inequalities
MJ
a

.MJ
b
MA.MB.MC

AJ
a
.AJ
b
bcMA
+
BJ
a
.BJ
b
caMB
+
CJ
a
.CJ
c
abMC
MJ
b
.MJ
c
MA.MB.MC

AJ
b
.AJ

c
bcMA
+
BJ
b
.BJ
c
caMB
+
CJ
b
.CJ
c
abMC
MJ
c
.MJ
a
MA.MB.MC

AJ
c
.AJ
a
bcMA
+
BJ
c
.BJ
a

caMB
+
CJ
c
.CJ
a
abMC
.
On adding the three inequalities, we find the inequality
MJ
a
.MJ
b
+ M J
b
.MJ
c
+ M J
c
.MJ
a
MA.MB.MC

AJ
a
.AJ
b
+ AJ
b
.AJ

c
+ AJ
c
.AJ
a
bcMA
+
BJ
a
.BJ
b
+ BJ
b
.BJ
c
+ BJ
c
.BJ
a
caMB
+
CJ
a
.CJ
b
+ CJ
b
.CJ
c
+ CJ

c
.CJ
a
abMC
.
Corollary 2.5. Let be the triangle ABC with the lengths of sides a, b, c and R is the
radius of circumcircle of ABC. Denote O, H the center of circumcircle and the or-
thocenter of ABC. Then, with any point M, we have the inequality:
abcMO.MH
RMA.M B.M C

aAH
MA
+
bBH
MB
+
c CH
MC
.
if M belongs to the circle with the centre O and the radius R, we obtain the inequality
abcMH
MA.MB.MC

a
√
4R
2
− a
2

MA
+
b
√
4R
2
− b
2
MB
+
c
√
4R
2
− c
2
MC
.
Proof. Applying the inequality (M, N ) to the case n = 3, s = 2 we have the inequality:
MO.MH
MA.MB.MC

AO.AH
bcMA
+
BO.BH
caMB
+
CO.CH
abMC

.
Thus, we obtain the inequality
abcMO.MH
RMA.M B.M C

aAH
MA
+
bBH
MB
+
c CH
MC
. Since AH =
√
4R
2
− a
2
, BH =
√
4R
2
− b
2
and CH =
√
4R
2
− c

2
we obtain
abcMH
MA.MB.MC

a
√
4R
2
− a
2
MA
+
b
√
4R
2
− b
2
MB
+
c
√
4R
2
− c
2
MC
.
19

Corollary 2.6. Suppose given the triangle ABC with the lengths of sides a, b, c, respec-
tively. I, G, H the center of inscribed circle and the barycenter and the orthocenter of
∆ABC. Then, with any point M, we always have the inequality
(i)
abcMI
2
MA.MB.MC

aAI
2
MA
+
bBI
2
MB
+
c CI
2
MC
.
(ii)
abcMG
2
MA.MB.MC

aAG
2
MA
+
bBG

2
MB
+
c CG
2
MC
.
(iii)
abcMH
2
MA.MB.MC

a(4R
2
− a
2
)
MA
+
b(4R
2
− b
2
)
MB
+
c(4R
2
− c
2

)
MC
.
Proof. Applying the inequality (M,N) to the case n = 3, s = 2, N
1
≡ N
2
≡ N, we have
MN
2
MA.MB.MC

AN
2
bcMA
+
BN
2
caMB
+
CN
2
abMC
.
Therefore, we obtain the inequality
abcMI
2
MA.MB.MC

aAI

2
MA
+
bBI
2
MB
+
c CI
2
MC
and
abcMG
2
MA.MB.MC

aAG
2
MA
+
bBG
2
MB
+
c CG
2
MC
. Then we have (i) and (ii). If N ≡ H we have (iii):
abcMH
2
MA.MB.MC


a(4R
2
− a
2
)
MA
+
b(4R
2
− b
2
)
MB
+
c(4R
2
− c
2
)
MC
.
Proposition 2.7. Suppose the polygon A
1
A
2
. . . A
n
is inscribed in the circle with the
center O and radius R. Then, w ith any s < n points N

1
, . . . , N
s
in the plane A
1
A
2
. . . A
n
,
we always have the inequality
n

k=1
s

i=1
A
k
N
i
n

i=1,i=k
A
k
A
i

s


i=1
ON
i
R
n−1
. When R = 1 we obtain
n

k=1
s

i=1
A
k
N
i
n

i=1,i=k
A
k
A
i

s

i=1
ON
i

.
When n = 3, s = 1 and a
1
= A
2
A
3
, a
2
= A
3
A
1
, a
3
= A
1
A
2
we obtain the inequality
a
1
A
1
N + a
2
A
2
N + a
3

A
3
N  4S
A
1
A
2
A
3
ON
R
.
Proof. The inequality
n

k=1
s

i=1
A
k
N
i
n

i=1,i=k
A
k
A
i


s

i=1
ON
i
R
n−1
follows from the inequality (M, N)
when M ≡ O. With R = 1 we have
n

k=1
s

i=1
A
k
M
i
n

i=1,i=k
A
k
A
i

s


i=1
OM
i
.
20
www.VNMATH.com
Example. Giving the triangle ABC with the lengths of sides a, b, c and R is the ra-
dius of circumscribed circle; r
1
, r
2
, r
3
. are the radius of escribed circles. Let’s d
a
, d
b
, d
c
the distances from the center of circumscribed circle to the center of escribed circles.
Then, with any point D belong to the circumscribed circle of ∆ABC we always have the
inequality:
(i)
d
a
d
b
d
c
√

a + b + c
R
3

√
bc
x
√
b + c − a
+
√
ca
y
√
c + a − b
+
√
ab
z
√
a + b − c
+
DJ
a
.DJ
b
.DJ
c
xyz
√

a + b + c
.
(ii)

(R + 2r
1
)(R + 2r
2
)(R + 2r
3
)
R
3
(a + b + c)

√
bc
√
b + c − a
+
√
ca
√
c + a − b
+
√
ab
√
a + b − c
+

DJ
a
.DJ
b
.DJ
c
xyz
√
a + b + c
.
Proof. (i) We consider M ≡ O.
Since













J
a
A
2
=

bc(a + b + c)
b + c − a
, J
a
B
2
=
ca(a + b − c)
b + c − a
, J
a
C
2
=
ab(a − b + c)
b + c − a
J
b
A
2
=
bc(b + a − c)
c + a − b
, J
b
B
2
=
ca(a + b + c)
c + a − b

, J
b
C
2
=
ab(b + c − a)
c + a − b
J
c
A
2
=
bc(c + a − b)
a + b − c
, J
c
B
2
=
ca(c + b − a)
a + b − c
, J
c
C
2
=
ab(a + b + c)
a + b − c
there-
fore we obtain

d
a
d
b
d
c
√
a + b + c
R
3

√
bc
x
√
b + c − a
+
√
ca
y
√
c + a − b
+
√
ab
z
√
a + b − c
+
DJ

a
.DJ
b
.DJ
c
xyz
√
a + b + c
.
(ii) Since d
2
a
= R
2
+ 2Rr
1
, d
2
b
= R
2
+ 2Rr
2
, d
2
c
= R
2
+ 2Rr
3

therefore

(R + 2r
1
)(R + 2r
2
)(R + 2r
3
)
R
3
(a + b + c)

√
bc
√
b + c − a
+
√
ca
√
c + a − b
+
√
ab
√
a + b − c
+
DJ
a

.DJ
b
.DJ
c
xyz
√
a + b + c
.
Now, we illustrate the advantage of this identity by addressing several important
problems of elementary Geometry. Firstly, we use the functions sin and cosin to create
the identity under the form of trigonometry.
Without generality, we can assume that the radius R of the circle C equal to 1. Suppose
that every point A
k
has affixe a
k
= cos α
k
+ i sin α
k
, and M has affixe z = cos u +
i sin u and every N
h
has affixe z
h
= cos u
h
+ i sin u
h
. Following the interpolation formula

21
Lagrange, we have
s

j=1
(z − z
j
)
n

t=1
(z − a
t
)
=
n

k=1
s

j=1
(a
k
− z
j
)
(z − a
k
)


t=k
(a
k
− a
t
)
or
s

j=1
2i sin
u − u
j
2
e
i(u + u
j
)
2
n

t=1
2i sin
u − α
t
2
e
i(u + α
t
)

2
=
n

k=1
s

j=1
2i sin
α
k
− u
j
2
e
i(α
k
+ u
j
)
2
2i sin
u − α
k
2
e
i(u + α
k
)
2


t=k
2i sin
α
k
− α
t
2
e
i(α
k
+ α
t
)
2
.
We reduce all the factors 2i, e
iu
j
2
and e
iα
t
2
. And we receive the relationship
s

j=1
sin
u − u

j
2
n

t=1
sin
u − α
t
2
=
n

k=1
s

j=1
sin
α
k
− u
j
2
sin
u − α
k
2

t=k
sin
α

k
− α
t
2
e
i(s + 1 − n)(α
k
− u)
2
.
From this relationship, we deduce 2 identities below:
s

j=1
sin
u − u
j
2
n

t=1
sin
u − α
t
2
=
n

k=1
s


j=1
sin
α
k
− u
j
2
sin
u − α
k
2

t=k
sin
α
k
− α
t
2
cos
(s + 1 − n)(α
k
− u)
2
and
n

k=1
s


j=1
sin
α
k
− u
j
2
sin
u − α
k
2

t=k
sin
α
k
− α
t
2
sin
(s + 1 − n)(α
k
− u)
2
= 0. With this result, we have
just built the identities under the form of trigonometry and geometry for the inequality
(M, N) as below:
Proposition 2.8. Assume that the polygon A
1

A
2
. . . A
n
is inscribed in the circle with
the center O and radius R. Taking s + 1 points N
1
, . . . , N
s
and M also belonging to this
circle C. Assuming that the coordinate A
k
(cos α
k
; sin α
k
), k = 1, 2, . . . , n; the coordinate
N
j
(cos u
j
; sin u
j
), j = 1, 2, . . . , s and the coordinate M(cos u; sin u). Then, we will have
these identities
(i)
s

j=1
sin

u − u
j
2
n

t=1
sin
u − α
t
2
=
n

k=1
s

j=1
sin
α
k
− u
j
2
sin
u − α
k
2

t=k
sin

α
k
− α
t
2
cos
(s + 1 − n)(α
k
− u)
2
22
www.VNMATH.com
(ii)
n

k=1
s

j=1
sin
α
k
− u
j
2
sin
u − α
k
2


t=k
sin
α
k
− α
t
2
sin
(s + 1 − n)(α
k
− u)
2
= 0.
(iii)
3

k=1
sin
α
k
− u
1
2

t=k
sin
α
k
− α
t

2
= 0 when n = 3, s = 1.
(iv)
n−1

j=1
sin
u − u
j
2
n

t=1
sin
u − α
t
2
=
n

k=1
n−1

j=1
sin
α
k
− u
j
2

sin
u − α
k
2

t=k
sin
α
k
− α
t
2
when s = n − 1.
(v)
n−2

j=1
sin
u
j
2
n

t=1
sin
α
t
2
=
n


k=1
n−2

j=1
sin
α
k
− u
j
2

t=k
sin
α
k
− α
t
2
cot
α
k
2
and
n

k=1
n−2

j=1

sin
α
k
− u
j
2

t=k
sin
α
k
− α
t
2
= 0 when s = n −2
and u = 0.
Remark 2.9. If the quadrilateral ABCD is inscribed in the circle we have
DA
bc
+
DC
ab
=
DB
ca
by (iii) or
aDA
2
DA
+

cDC
2
DC
=
bDB
2
DB
. Using the relationship
DA
2
.DB.DC.a + DC
2
.DA.DB.c = DB
2
.DC.DA.b.
Hence DA
2
S
DBC
+ DC
2
S
DAB
= DB
2
S
DCA
[Feuerbach].
Proposition 2.10. Suppose the polygon A
1

A
2
. . . A
n
is inscribed in the circle with
the radius R = 1. Taking s + 1 points N
1
, . . . , N
s
and M also belonging to this cir-
cle C. Assuming that the coordinate A
k
(cos α
k
; sin α
k
), k = 1, 2, . . . , n; the coordinate
N
j
(cos u
j
; sin u
j
), j = 1, 2, . . . , s and the coordinate M(cos u; sin u). Then, with the proper
choices of + or − we will have the identities
(i)
s

j=1
MN

j
n

t=1
MA
t
=
n

k=1
±
s

j=1
A
k
N
j
MA
k

t=k
A
k
A
t
cos
(s + 1 − n)(α
k
− u)

2
, (M, N).
(ii)
n

k=1
±
s

j=1
A
k
N
j
MA
k

t=k
A
k
A
t
sin
(s + 1 − n)(α
k
− u)
2
= 0.
23
(iii)

n−2

j=1
MN
j
n

t=1
MA
t
=
n

k=1
±
n−2

j=1
A
k
N
j

t=k
A
k
A
t
cot
α

k
2
and
n

k=1
±
n−2

j=1
A
k
A
j

t=k
A
k
A
t
= 0.
Corollary 2.11. Asuming that the points A
1
, A
2
, . . . , A
n
, M in order belong to the circle
C with the center O. We convent n + 1 := 1. Then, we have the identities
(i)

n

r=1
(−1)
r
cos(n − 1)∠MA
r+1
A
r
MA
r

k=r
A
r
A
k
=
1
n

k=1
MA
k
.
(ii)
n

r=1
(−1)

r
sin(n − 1)∠MA
r+1
A
r
MA
r

k=r
A
r
A
k
= 0.
Proof. These identities follow from the identity (M, N ) with s = 0
Corollary 2.12. Let the quadrilateral ABCD be inscribed in the circle C with the center
O. Let a = BC, b = CA, c = AB. Then, we have 2 identities:
(i)
a cos(OD, OA)
DA
−
b cos(OD, OB)
DB
+
c cos(OD, OC)
DC
= −
abc
DA.DB.DC
.

(ii)
a sin(OD, OA)
DA
+
c sin(OD, OC)
DC
=
b sin(OD, OB)
DB
.
Proof. These identities follow from the identity (M, N ) with n = 3, s = 0.
3 Conjecture
Despite of not having been proven yet, these following results are still hoped to be true:
Open Problem 3.1. Suppose given a triangle ABC with the lengths of sides a, b, c
respectively and R is the radius of circumcircle of ABC. Let’s I, J
a
, J
b
, J
c
are the
centres of incircle and escribed circles of ∆ABC, respectively. Then, with any point M,
we have
(i)
MJ
a
.MJ
b
.MJ
c

MA.MB.MC

AJ
a
.AJ
b
.AJ
c
bcMA
+
BJ
a
.BJ
b
.BJ
c
caMB
+
CJ
a
.CJ
b
.CJ
c
abMC
.
(ii)
MJ
a
.MJ

b
.MJ
c
√
a + b + c
MA.MB.MC

√
bc
√
b + c − a
MA
+
√
ca
√
c + a − b
MB
+
√
ab
√
a + b − c
MC
.
Open Problem 3.2. Giving the triangle ABC with the lengths of sides a, b, c and R
is the radius of circumscribed circle; r
1
, r
2

, r
3
. are the radius of escribed circles. Let’s
d
a
, d
b
, d
c
the distances from the center of circumscribed circle to the center of escribed
circles. Then, with any point D belong to the circumscribed circle of ∆ABC we always
have the inequality:
24
www.VNMATH.com
(i)
d
a
d
b
d
c
√
a + b + c
R
3

√
bc
x
√

b + c − a
+
√
ca
y
√
c + a − b
+
√
ab
z
√
a + b − c
.
(ii)

(R + 2r
1
)(R + 2r
2
)(R + 2r
3
)
R(a + b + c)

√
bc
√
b + c − a
+

√
ca
√
c + a − b
+
√
ab
√
a + b − c
.
References
[1] T. Andreescu and D. Andrica, Proving some geometric inequalities by using complex
numbers, Eduatia Mathematica Vol 1. Nr. 2 (2005),19-26.
[2] T. Hayashi, Two theorems on complex numbers, Thoku Math. J., 4(1913/14), pp.
68-70.
[3] D.S. Mitrinovic, J.E. Pecaric and V. Volenec, Recent Advances in Geometric In-
equalities, Acad. Publ., Dordrecht, Boston, London 1989.
25