Extensions of a category of inequalities

Vol. 17, No.2, October 2009, pp 639-647 ISSN 1222-5657, ISBN 978-973-88255-5-0, www.hetfalu.ro/octogon

639

Extensions of a category of inequalities

D.M. B˘atinet¸u-Giurgiu and Mih´aly Bencze¹²

ABSTRACT.In this paper we present extensions of a category of inequalities, which have applications in fundamental theory of inequalities, and evidently in mathematical contests.

MAIN RESULTS

Theorem 1. If a1, ..., ak, b1, ..., bp, xt, yt>0, α, γ ∈R,

β∈(−∞,0]∪[2,+∞) and i1, ..., ik, j1, ..., jp∈ {1,2, ..., n} are different indexes, then

X

cyclic

¡a1x^α_i1+...+akx^α_ik¢_β b1y^γ_j1+...+bpy_j^γ_p ≥

n^2−β(a1+...+ak)^p µ _n

P

t=1

x^α_t

¶_β

(b₁+...+b_p)Pⁿ

t=1

y_t^γ Proof. Using first the Cauchy-Schwarz‘s and after them the Jensen‘s inequality we obtain:

(b1+...+bp) Xn

t=1

y^γ_t X

cyclic

¡a1x^α_i1+...+a_kx^α_ik¢_β b₁y_j^γ₁+...+b_py^γ_jp =

=



X

cyclic

³

b1y_j^γ₁+...+bpy_j^γ_p

´

 X

cyclic

¡a1x^α_i1+...+akx^α_ik¢_β b1y_j^γ₁+...+bpy^γ_jp ≥

≥



X

cyclic

¡a1x^α_i1+...+akx^α_ik¢^β

2





2

≥

12Received: 25.03.2006

2000Mathematics Subject Classification. 26D15

Key words and phrases. Cauchy‘s inequality, Jensen‘s inequality

≥



 1 n^β2−1

X

cyclic

¡a₁x^α_i1 +...+a_kx^α_ik¢^β

2





2

=n^2−β(a₁+...+a_k)^β Ã _n

X

t=1

x^α_t

!_β

Corollary 1.1. If a, b, x_t>0 (t= 1,2, ..., n) andβ ∈(−∞,0]∪[2,+∞), then

X

cyclic

x^β₁ ax₂+bx₃ ≥

n^2−β µPn

t=1

x^α_t

¶_β

a+b

Proof. In Theorem 1 we takeα= 1, a₁= 1, a₂ =a₃ =...=a_k= 0, b₁=a, b₂ =b, y₁ =x₂, y₂ =x₃, b₃=b₄=...=b_p = 0, γ = 1.

Ifa=b= 1 then we obtain the Corollary 2.1 from [2].

If Qⁿ

t=1

x_t= 1,then Pⁿ

t=1

x_t≥nand we get X

cyclic

x^β₁

ax₂+bx₃ ≥ n a+b which is the Remark 2.1.2 from [2].

Ifx_t→ _x¹

t (t= 1,2, ..., n),then we get X

cyclic

x₂x₃

x^β₁(ax₃+bx₂) ≥ n^2−β

µ _n P

t=1 x1t

¶_β

a+b If Qⁿ

t=1

x_t= 1,then from the previous inequality we obtain X

cyclic

x₂x₃

x^β₁ (ax₂+bx₃) ≥ n a+b Ifn= 3,β = 2 and x₁x₂x₃= 1,then

X 1

x³₁(ax₂+bx₃) ≥ 3 a+b

is a problem proposed by Russia in 1995 and presented at IMO.

Corollary 1.2. If a, b, x_t>0 (t= 1,2, ..., n),(a≥b) and β ∈(−∞,0]∪[2,+∞), then

X

cyclic

x^β₁

a(x₁+...+x_k−1) + (a−b)x_k+a(x_k+1+...+x_n) ≥ n^2−β

µ _n P

t=1

x_t

¶_β−1

an−b Proof. In Theorem 1 we takea₁ = 1, a₂=...=a_k= 0, x_t=y_t

(t= 1,2, ..., n), b₁=...=b_k−1 =b_k+1=...=b_n=a, b_k=a−b, α=γ = 1.

Corollary 1.3. If a, b, c, d, x_t>0 (t= 1,2, ..., n), β ∈(−∞,0]∪[2,+∞), then

X

cyclic

(ax₁+bx₂)^β xc₃+dx₄ ≥

n^2−β(a+b)^β µ _n

P

t=1

x_t

¶_β

c+d from this we obtain

X

cyclic

(x₁+x₂)^β

x₃+u ≥ 2^βn^2−β nu+ 1 which is Corollary 2.4.1 from [2].

Ifβ = 2, then we get

X

cyclic

(x₁+x₂)^β

x₃+u ≥ 4

1 +nu

Ifu= 1 and n= 3 then we obtain the problem 24380 from Gazeta

Matematica 10/2000, author Mihai Opincariu, namely ifa+b+c= 1, then X(a+b)²

c+ 1 ≥1

After some modification we get the following: Ifa, b, c >0 then X (a+b)²

a+b+ 2c ≥a+b+c

In same way we obtain the following:

X

cyclic

x^β₁

x₂+x₃ ≥ 1 2n^2−β

à _n X

t=1

x_t

!_β−1

which is Corollary 2.6 from [2].

Ifβ = 2,then we get the following: Ifa, b, c, d >0 anda+b+c+d= 1, then X a²

a+b ≥ 1 2

which is a problem presented by Ireland at IMO 1999.

Corollary 1.4. If b₁, ..., b_p, x_t>0 (t= 1,2, ..., n), β ∈(−∞,0]∪[2,+∞), then

X

cyclic

(x₁+x₂+...+x_k)^β b₁x₁+b₂x₂+...+b_px_p ≥

k^β µ _n

P

t=1

x_t

¶_β−1

(b₁+...+b_p)n^β−2 which is Corollary 2.7 from [2].

Ifk= 2, β= 2, b₁ =b₂ = 1, b₃ =b, b₄=...=b_p = 0,then we get the following: If x, y, z, b >0, then

X (x+y)²

x+y+bz ≥ 4 (x+y+z) b+ 2

which is problem C.2403 from Gazeta Matematica 5-6/2001, author Titu Zvonaru.

After some substitutions we get:

X 1

x₁+x₂+...+x_k ≥ n² kPⁿ

t=1

x_t

Ifn= 3, k= 2 then we obtain the following: Ifa, b, c >0, then X 1

a+b ≥ 9

2 (a+b+c)

which was a problem in Ireland at a Mathematical Contest in 1999.

After another specification we obtain the following:

Corollary 1.5. If x_t> u >0 (t= 1,2, ..., n) andβ ∈(−∞,0]∪[2,+∞), then

X x^β₁ x₂−u ≥

µ _n P

t=1

x_t

¶_β

n^β−2 µ _n

P

t=1

x_t−nu

¶

Ifβ = 2, a, b≥u= 1, n= 2, then we obtain the following: If a, b≥1, then a²

b−1+ b² a−1 ≥8

which was a problem in Russian Mathematical Olympiad 1992. From Corollary 1.5 we get:

a²

b−1 + b²

a−1 ≥ (a+b)² a+b−2 but (a+b)² ≥8 (a+b−2)⇔(a+b−4)²≥0.

In same way we obtain the following problem: Ifa, b≥2, then a²

b−2 + b²

a−2 ≥16

which was presented at Mathematical Olympiad in Moldavia. From Corollary 1.5 we get

a²

b−2 + b²

a−2 ≥ (a+b)² a+b−4 but

(a+b)²≥16 (a+b−4)⇔(a+b−8)² ≥0 Corollary 1.6. If a, b, x_t>0 (t= 1,2, ..., n),then

X

1≤i<j≤n

1 ax_ix_j +b ≥

¡_n

2

¢₂

a P

1≤i<j≤n

x_ix_j+¡_n

2

¢b ≥

¡_n

2

¢₂

a(n−1) 2

Pn t=1

x²_t+¡_n

2

¢b

which is Corollary 2.10.1 from [2].

If Pⁿ

t=1

x²_t =n, then

X

1≤i<j≤n

1

ax_ix_j+b ≥ n(n−1) 2 (a+b) Ifa= 1 then we obtain corollary 2.10.2 from [2].

Ifn= 3,a=b= 1 then we obtain the following: Ifa, b, c >0 and a²+b²+c² = 3,then

X 1

ab+ 1 ≥ 3 2

which is a problem presented at Mathematical Olympiad in Belarus 1999.

The result of corollary 1.6 can be generalized in following way: Ifa, b, x_t>0 (t= 1,2, ..., n) then

X

1≤i1<...<ik≤n

1

ax_i1x_i2...x_ik+b ≥

¡_n

k

¢₂

a P

1≤i1<...<ik≤n

x_i1...x_ik+¡_n

k

¢b

Corollary 1.7. If b_t, x_t>0 (t= 1,2, ..., n) and β ∈(−∞,0]∪[2,+∞),then x^β₁

b1

x1 +_x^b2

2 +...+_x^bn

n

+ x^β₂

b1

x2 +_x^b2

3 +...+ ^b_xⁿ

n

+...+ x^β_n

b1

xn + ^b_x²

1 +...+_x^bn

n

≥

≥

µ _n P

t=1

x_t

¶_β

n^β−1 µ _n

P

t=1 x1t

¶ µ _n P

t=1

b_t

¶

Proof. In Theorem 1 we takea₁ = 1, a₂ =a₃=...=a_k= 0, p=n, y_t= _x¹_t (t= 1,2, ..., n).

After some particularization we get the following problem: If a, b, c, d, e >0 and β∈(−∞,0]∪[2,+∞),then

X a^β

b+c ≥ (P a)^β−1 2·5^β−2 Ifm= 1 then we get

X a b+c ≥ 5

2

which is a problem of selection test in USA, for IMO 2001, author Titu Andreescu.

Corollary 1.8. If a, b, c >0 andβ ∈(−∞,0]∪[2,+∞),then X a^β

a²−bc+ 1 ≥ 3^2−β(P a)^β (P

a)²−3P ab+ 3 IfP

ab= ¹₃,then

X a^β

a²−bc+ 1 ≥3^1−β³X a

´_β−2

Ifβ = 1, then

X a

a²−bc+ 1 ≥ 1 Pa

which is problem O:976 Gazeta Matematica 9-10/2001, author Titu Andreescu.

Corollary 1.9. If a, x, y, z, t >0 and β ∈(−∞,0]∪[2,+∞),then X x^β

x³−3yzt+ax ≥ 4^2−β(P x)^β−1 Px²−P

xy+a Ifβ = 2,then

X x²

x³−3yzt+ax ≥

Px Px²−P

xy+a IfP

xy = ^a₃,then

X x^β

x³−3yzt+a ≥4^2−β³X x

´_β−3

Corollary 1.10. Ifx_t, y_t>0 (t= 1,2, ..., n) andk∈N,then

ky₁^k+1+ Xn

t=2

y_t^k+1 x_t−1 ≥

(k+ 1)y^k₁ Pⁿ

t=2

y_t

k+1

s

n^k−1 Pⁿ

t=2

x_t−1

Proof. Using the Theorem 1 we get

ky₁^k+1+ Xn

t=2

y_t^k+1

x_t−1 ≥ky₁^k+1+

µPn

t=2

y_t

¶_k+1

(n−1)^k−1 Pⁿ

t=2

x_t−1

=y₁^k+1+...+y₁^k+1

| {z }

k−time

+

+

µPn

t=2

y_t

¶_k+1

(n−1)^k−1Pⁿ

t=2

x_t−1

≥

(k+ 1)y^k₁ Pⁿ

t=2

y_t

k+1

s

(n−1)^k−1Pⁿ

t=2

x_t−1 If Pⁿ

t=2

x_t−1= 1,then we get

ky₁^k+1+ Xn

t=2

y_t^k+1 x_t−1 ≥

(k+ 1)y^k₁ Pⁿ

t=2

y_t

k+1

q

(n−1)^k−1 which is Corollary 2.11.1 from [2].

Ifk= 1 then we get y₁²+ y₂²

x₁ + y₃²

x₂ +...+ y_n²

x_n−1 ≥2y₁ Xn

t=2

y_t

which is a problem from O.B.M.J, 2000 author D.Acu.

Corollary 1.11. Ifx_t>0 (k= 1,2, ..., n),and y₀ > y₁ > ... > y_n, then

ky₀+ Xn

t=1

x^k+1_t

y_k−1−y_k ≥(k+ 1) Ã _n

X

t=1

x_t

!

k+1

sµy₀−y_n n

¶_k−1 +ky_n Ifx_t= 1 (t= 1,2, ..., n) and k= 1 then we obtain:

y₀+ Xn

t=1

1

y_t−1−y_t ≥y_n+ 2n

which is a problem presented at Mathematical Olympiad Sankt Petersburg 1999.

REFERENCES

[1] B˘atinet¸u-Giurgiu, D.M.,Asupra unei probleme de la O.I.M., Canada 1995, Revista de Matematica si de Informatica din Constanta, Anul IX, Nr.

2/2000, pp. 3-4.

[2] Bencze, M. and Chang-Jian, Z.,Applications of generalized

Cauchy-Bunjakowsky-Schwarz inequality, Octogon Mathematical Magazine, Vol. 10, No. 2, 2000, pp. 591-598.

[3] Octogon Mathematical Magazine (1993-2009) Calea 13 Septembrie 59-61,

Bl 59-61, Sc. 2, Ap. 28 05071, Bucuresti, Romania Str. H˘armanului 6,

505600 S˘acele-N´egyfalu Jud. Bra¸sov, Romania

E-mail: benczemihaly@yahoo.com