• Nu S-Au Găsit Rezultate

# Several inequalities of Erdős-Mordell type

N/A
N/A
Protected

Academic year: 2022

Share "Several inequalities of Erdős-Mordell type"

Copied!
21
0
0

Text complet

(1)

### Several inequalities of Erdős-Mordell type

Mihály Bencze and Marius Drăgan

Str. Hărmanului 6, 505600 Săcele-Négyfalu, Jud. Braşov, Romania E-mail: [email protected]

61311 bd. Timişoara Nr. 35, Bl. 0D6, Sc. E, et. 7, Ap. 176, Sect. 6, Bucureşti, Romania E-mail: [email protected]

ABSTRACT. The purpose of this paper is to obtain some inequalities between sum or product of power to order,k∈[0,1]relate to distences from an interior pointM of triangleABC to the vertices of triangle, radii of triangleM BC, M CA, M ABand distances from M to the sidesBC, CA, AB and using this inequalities and an inversion of polM and ratiotin conexion with verticesA, B, Cto obtain many other new inequalities.

1 INTRODUCTION

Let be M an interior point of triangle ABC,Ra, Rb, Rc the radii of the circumscribe of triangles

M BC, M CA, M AB, R1, R2, R3the distances fromM to the verticesA, B, C andr1, r2, r3 the distances fromM to the sidesBC, CA, AB.

The inequality Erdős-Mordell

R1+R2+R32 (r1+r2+r3) (1.1) is true in any triangle ABC.

Also a generalization of Erdős-Mordell appears in [1] who said that:

x2R1+y2R2+z2R32 (yzr1+zxr2+xyr3) (1.2) is true for any real numbersx, y, z≥0.

If we takeλ1=x2, λ2=y2, λ3=z2 are true inequality λ1R1+λ2R2+λ3R32(√

λ2λ3r1+√

λ3λ1r2+√ λ1λ2r3

)

(1.3)

who appears in [2].

Also

λ21Ra+λ22Rb+λ23Rc≥λ1λ2λ3

(R1

λ1 +R2

λ2 +R3

λ3 )

(1.4)

Who are published in [3].

In the following we give a lot of inequalities of this type and many other in connection with them.

Lemma 1. Let beM an interior point of triangle ABC.Then we have:

a). 1ac ·2RR3b +ba·2RR2c b). 1ab ·2RR3b +ac ·2RR2c

Proof.

a). We use the well-known inequality

aR1≥cr2+br3

and the equality

r1= R2R3

2Ra who result writing the area of triangle MBC

(2)

SABC =R2R3

4Ra = ar1

2 b). It result from the inequality

aR1≥br2+cr3

Lemma 2. Let beM an interior point of triangle ABC.Then we have:

R2 Rc

+R3

Rb 4 sinA 2

We considerU =prABM , V =prMAC. u=µ(∡U AM), v=µ(∡V AM).We have r2+r3=R1(sinu+ sinv) = 2R1sinA

2 cosu−v

2 2R1sinA 2 or

R3R1

2Rb +R1R2

2Rc 2R1sinA 2 Who is equivalent with the inequality from the statement.

Theorem 1. Let beM a interior point of triangle ABC. Then for every real numberk∈[0,1], λ1, λ2, λ30.We have:

a). λ1Rka+λ2Rkb+λ3Rkc ≥√

λ2λ3Rk1+

λ3λ1R2k+ λ1λ2Rk3 b).

λ2λ3RkRk1

b+Rkc +

λ3λ1RkRk2

c+Rak+

λ1λ2RkRk3

a+Rkb

12 (

λ2λ3 Rk1

RkbRkc + λ3λ1

Rk2

RkcRka + λ1λ2

Rk3

RkaRkb

)

λ1223 Proof. a). From the inequality(x+y)k2k1(

xk+yk)

, k∈[0,1]it result that:

(a c · R2

2Ra +b c · R1

2Rb )k

2k1 ([(a

c

)k Rk2 (2Ra)k +

(b c

)k( R1

2Rb )k])

Using this inequality and Lemma 1. We obtain

1 (a

c · R2

2Ra +b R1

2Rb )k

2k1 [(a

c

)k Rk2 (2Ra)k +

(b c

)k( Rk1 2Rb

)k]

or

2≥ak ck ·R2k

Rak +bk ck ·Rk1

Rkb or

3Rkc ≥λ3

ak ck · Rkc

Rka ·Rk2+λ3

bk ck ·Rkc

Rkb ·Rk1 and the similar inequalities

2Rkb ≥λ2ck bk ·Rkb

Rkc ·Rk1+λ2ak bk ·Rkb

Rka ·Rk3

1Rka ≥λ1bk sk ·Rka

Rkb ·Rk3+λ1ck ak · Rka

Rkc ·Rk2 Adding this inequalities and apply the A-M inequalities we obtain:

2(

λ1Rka+λ2Rbk+λ3Rkc)

≥Rk1 (

λ3·bk ck · Rkc

Rkb +λ2· ck bk · Rkb

Rkc )

+

(3)

+R2k (

λ3·ak ck ·Rkc

Rka +λ1· ck ak · Rka

Rkc )

+Rk3 (

λ2· ak bk ·Rkb

Rka +λ1· bk ak ·Rka

Rkb )

2√

λ2λ3R1k+ 2√

λ3λ1Rk2+ 2√ λ2λ1Rk3

b). Adding the inequalities

3≥λ3·ak ck · Rk2

Rka +λ3· bk ck ·Rk1

Rkb

1≥λ1· bk ak ·Rk3

Rkb +λ1· ck ak · Rk2

Rkc

3≥λ2·ck bk ·Rk1

Rc+λ2·ak bk ·Rk3

Rka We obtain

2 (λ1+λ2+λ3) (

λ3·ak ck · Rk2

Rka +λ1· ck ak ·Rk2

Rkc )

+

+ (

λ3·bk ck ·Rk1

Rkb +λ2·ck bk ·Rk1

Rkc )

+ (

λ1· bk ak ·Rk3

Rkb +λ2· ak bk ·Rk3

Rka )

2√

λ1λ3· Rk2

RkaRkc + 2√

λ2λ3· Rk1

RkbRkc

+ 2√

λ1λ2· Rk3

RkaRkb

4√

λ1λ3· Rk2

Rka+Rkc + 4√

λ2λ3· Rk1

Rkb+Rkc + 4√

λ1λ2· Rk3 Rak+Rkb

We observe that b) imply a)

Indeed putting in b). λ1→λ1Rka, λ2→λ2Rkb, λ3→λ3Rkc,we obtain

RkbRck· Rk1·√ λ2λ3

Rbk+Rck +

RkcRka· Rk2·√ λ3λ1

Rka+Rkc +

RkaRkb ·Rk3·√ λ1λ2

Rka+Rkb

1 2

(λ1Rk1+λ2Rk2+λ3Rk3)

1 2

(λ1Rak+λ2Rkb +λ3Rck)

Also putting in b). λ1→λ1Rkb, λ2→λ2Rkc, λ3→λ3Rka andλ1→λ2Rkb, λ2→λ3Rkc, λ3→λ1Rka,we obtain:

Theorem 2. Let beM an interior point of triangleABC. Then for ecery real numberk∈[0,1], λ1, λ2, λ30, we have:

a).

λ2λ3·Rk1Rkc ·Rka Rkb+Rkc +√

λ3λ1· Rk2

Rka·Rkb Rka+Rkc +√

λ1λ2·Rk3

Rkb ·Rkc Rka+Rkb

1 2

√ λ2λ3Rk1

Rka Rkb +√

λ3λ1Rk2

Rkb Rkc +√

λ1λ2R3k

Rkc Rka

1 2

(λ1Rkb +λ2Rkc+λ3Rka) b).

(4)

λ3λ1·R1kRkc ·Rka Rkb+Rkc +√

λ1λ2· Rk2

Rka·Rbk Rkc +Rka +√

λ2λ3·Rk3

Rbk·Rkc Rka+Rkb

1 2

√ λ3λ1Rk1

Rka Rkb +√

λ1λ2Rk2

Rkb Rkc +√

λ2λ3Rk3

Rkc Rka

1 2

(λ2Rkb +λ3Rkc+λ1Rka)

If in Theorem 1 and 2 we takeλ1=λ2=λ3 we obtain

Corollary 2.1. a).

Rka+Rkb +Rkc ≥Rk1+Rk2+Rk3 b).

Rk1

Rkb +Rkc + Rk2

Rkc +Rka + Rk3

Rka+Rkb 1 2

Rk1

RkbRkc

+ Rk2

RkcRka + R3k

RkaRkb

3 2

c).

Rk1RkcRka Rkb +Rak +

Rk2

RkaRbk Rka+Rkc +

Rk3

RkbRkc Rka+Rkb 1

2

Rk1

Rka Rkb +Rk2

Rkb Rkc +Rk3

Rkc Rka

1 2

(Rka+Rkb +Rkc)

If we take k= 1in a). we obtain the following inequality

Ra+Rb+Rc≥R1+R2+R3

who appears in [1].

Also if we takek= 1 in b). we obtain:

R1

Rb+Rc + R2

Rc+Ra + R3

Ra+Rb 1 2

( R1

√RbRc

+ R2

√RaRb

+ R3

√RaRb

)

who represent a refinement of inequality RR1

b+Rc +RR2

c+Ra +RR3

a+Rb 32 who apperas in [2].

If we take in c). k= 1we obtain R1

√RcRa

Rb+Rc +R2

√RaRb

Ra+Rc +R3

√RbRc

Ra+Rc 1 2

( R1

Ra

Rb +R2

Rb

Rc +R3

Rc

Ra )

1

2(Ra+Rb+Rc) If in Theorem 2) we takeλ1→λ21, λ2→λ22, λ3→λ23,we obtain

Corollary 2.2. Let beM an interior point of triangleABC. Then for every real numberk∈[0,1], λ1, λ2, λ30, we have

λ21Rka+λ22Rkb +λ23R2c ≥λ1λ2λ3

(Rk1 λ1 +Rk2

λ2 +Rk3 λ3

) If we take k= 1we obtain

(5)

λ21Ra+λ22Rb+λ23Rc≥λ1λ2λ3

(R1

λ1 +R2

λ2 +R3

λ3 )

who represent just inequality (1.4).

Corollary 2.3. Let be M an interior point of triangleABC.Then for every real numberk∈[0,1]we have

a).

Rk1Rka+Rk2Rkb +Rk3Rkc ≤RkaRbk+RkbRkc +RckRka b).

1 + Rkb Rkc +Rkc

Rkb Rk1

RkbRkc

+ Rk2

RkaRkb

+ Rk3

RkaRkc

and similar inequalities c).

Rk1Rka+R2k

RkbRkc +Rk3

RkbRkc ≤RkaRkb +RkbRkc +RkcRkb and similar inequalities

d).

Rka Rkb +Rkb

Rkc +Rck

Rak Rk1

RkcRka + Rk2

RkbRka

+ Rk3

RkbRck

and similar inequalities e).

Rk1

RakRkb +Rk2

RkbRkc +Rk3

RkaRkc ≤RkaRkb +RkbRkc +RkcRkb Proof. We take in theorem 1

1, λ2, λ3) = ( 1

Rak, 1 Rbk, 1

Rck )

,1, λ2, λ3) = ( 1

Rka, 1 Rkb, 1

Rkc )

1, λ2, λ3) = ( 1

Rbk, 1 Rak, 1

Rck )

,1, λ2, λ3) = ( 1

Rkb, 1 Rkc, 1

Rka )

1, λ2, λ3) = ( 1

Rck, 1 Rak, 1

Rbk )

,1, λ2, λ3) = ( 1

Rkc, 1 Rkb, 1

Rka )

Corollary 2.4. Let be M an interior point of triangleABC.From every real number k∈[0,1],we have:

a).

Rk1 Rkb +R2k

Rck +Rk3 Rka ≤Rbk

Rck +Rkc Rka +Rka

Rkb

b).

Rk1 Rkb

Rka Rkc +Rk2

Rkc

Rkb Rka +Rk3

Rka

Rkc Rkb ≤Rbk

Rak +Rkc Rkb +Rka

Rkc

c).

(6)

Rk1 Rkb +R2k

Rck

Rkb

Rka + Rk3

RkbRka

≤Rkb Rkc +Rkc

Rkb + 1

and similar inequalities d).

Rk1

RkbRkc

+ Rk2

RkcRka + Rk3

RkaRkb

3

Proof. We take in Theorem 2:

1, λ2, λ3) = ( 1

Rak, 1 Rkb, 1

Rkc )

,1, λ2, λ3) = ( 1

Rka, 1 Rkb, 1

Rkc )

1, λ2, λ3) = ( 1

Rbk, 1 Rka, 1

Rkc )

,1, λ2, λ3) = ( 1

Rkb, 1 Rkc, 1

Rka )

1, λ2, λ3) = ( 1

Rck, 1 Rka, 1

Rkb )

,1, λ2, λ3) = ( 1

Rkc, 1 Rkb, 1

Rka )

Corollary 2.5. Let be M an interior point of triangleABC. Then for every real numberk∈[0,1]we have

a).

R12kRkc +R2k2 Rka+R2k3 Rkb ≥R2k1 R3k+R2k2 Rk1+R32kRk2 b).

R2k1 Rk3

RkbRck

+ R2k2 Rk1

RkcRka + R2k3 Rk2

RakRkb

≤R2k1 +R2k2 +R2k3

c).

R12kRk3

Rka

Rkb +R2k2 R1k

Rkb

Rkc +R2k3 Rk2

Rkc

Rka ≤R2k2 Rbk+R2k3 Rkc +R12kRka d).

R2k1 Rk2

Rka

Rkb +R2k2 Rk3

Rkb

Rkc +R2k3 R1

Rkc

Rka ≤R2k3 Rkb +R2k1 Rkc +R2k2 Rka Proof. We take

1, λ2, λ3) = ( Rk2

Rk1Rk3, Rk3 Rk1Rk2, R1k

Rk2Rk3, )

in Theorem 1 and 2

Corollary 2.6. Let be M an interior point of triangleABC. Then for every real numberk∈[0,1]we have

a).

Rka R2k2 + Rkb

R2k3 + Rkc R2k1 1

R1k + 1 Rk2 + 1

Rk3 b).

(7)

1

Rk3

RkbRkc

+ 1

Rk1

RkcRka + 1 R2k

RkaRkb

1 R2k1 + 1

R2k2 + 1 R2k3

Proof. We take in Theorem 1:

1, λ2, λ3) = ( 1

R2k2 , 1 R2k3 , 1

R2k1 )

Theorem 3. Let beM a interior point of triangleABC. Then for every real numberk∈[0,1]we have a).

RaRbRc

R1R2R3 1 81k ·

∏ (bk+ck)1/k

abc 1 b).

∏ (Rk3 Rkb +Rk2

Rkc )

64 (abc)k

∏(bk+ck) 8

Proof. From the inequalitiesR1 car2+abr3, R1 abr2+car3 and(x+y)k2k1(

xk+yk) ,if k∈[0,1]it result.

Rk1 (b

ar2+ c ar3

)k

2k1 (bk

akrk2+ck akr3k

)

and

Rk1 (c

ar2+ b ar3

)k

2k1 (ck

akrk2+bk akr3k

)

Adding the last two inequalities we obtain

2Rk1 2k1 [

rk2

(bk+ck ak

) +rk3

(bk+ck ak

)]

= 2k1(

r2k+rk3) (

bk+ck) ak

and the similar inequalities.

By multiplication we obtain:

8 (R1R2R3)k8k1

∏ (r2k+rk3) (

bk+ck) akbkck Taking account by equalitiesr2=R2R3R1

b , r3= R2R1R2

c we obtain:

8 (R1R2R3)k8k1∏ (Rk3R1k

2kRbk +Rk1Rk2 2kRkc

) ∏ (

bk+ck) akbkck or

64∏ (Rk3 Rkb +R2k

Rck ) ∏ (

bk+ck)

akbkck 8∏ (Rk3 Rkb +Rk2

Rkc )

or

∏ (Rk3 Rkb +Rk2

Rkc )

64akbkck

∏(bk+ck) 8

b). We have

64∏ (Rk3 Rkb +Rk2

Rkc ) ∏ (

bk+ck) akbkck 8

∏ (bk+ck)

akbkck · (R1R2R3)k (RaRbRc)k.

Referințe

DOCUMENTE SIMILARE

But let me start by stating in a preliminary way that by ‘The Muslim Question’ — the name is derived from an analogy with the eastern Question which be- devilled statesmen of the

Lactobacillus plantarum BLN39 showed better bacteriocin production against M. fortuitum MTCC1902 in the MRS broth medium supplemented with molasses. Molasses proved to be an

In this paper, a weighted logarithmic barrier interior-point method for solving the linearly convex constrained optimization problems is presented.. Unlike the classical

In section 4, we deal with the numerical implementation of the infeasible interior point algorithm applied to convex quadratic and linear problems.. We conclude that (QP ) µ and (QD)

Semidefinite linear complementarity problems, interior point meth- ods, long and small-update primal-dual algorithms, polynomial complexity.. Its growing importance can be measured

, Midpoint-Type Rules from an Inequalities Point of View, Handbook of Analytic-Computational Methods in Applied Mathematics, Editor:.. Anastassiou, CRC Press, New

S., Midpoint-type Rules from an Inequalities Point of View, Handbook of Analytic-Computational Methods in Applied Mathematics, Editor:.. Anastassiou, CRC Press, New

Goldman has defined in [3] an approximation operator of Stancu type for which has proved that it preserves the first degree functions.. In this paper, we begin the study of the

These trajectories, starting in a neighborhood of an asymptotic stable point, tend to the boundary of stability region, while the trajectories, starting near an equilibrium point on

Most of the results on infeasible-interior-point algorithms have been obta- ined for linear programming. The best computational complexity results obtained so far show

(VII) (PASCH) Si une droite qui ne passe pas par les sommets d’un triangle coupe un cˆ ot´ e du triangle (transversale), alors elle coupe encore au moins un des deux

The Wiener index of a graph is denoted by and is defined by .In general this kind of index is called a topological index, which is a distance based quantity assigned to

There are 4·2n −1 ways to choose the vertex of degree one, and the first i+2 vertices (including the ﬁrst vertex chosen for the reverse direction) are uniquely determined once

The molecular graph of NS 1 [n] has three similar branches with the same number x′ 23 of edges connecting a vertex of degree 2 with a vertex of degree 3.. Suppose y 23 is the

In the paper, by virtue of the H¨ older integral inequality, the authors derive some inequalities of the Tur´ an type for confluent hypergeometric functions of the second kind, for

According to our previous investigations it seems that tolerance, whether regarded as a political practice or a philosophical or moral principle, is a strategy (or tactics) of one

This can be done by replacing each axis with a vector of length one that points in the positive direction of the axis. Let O be the origin of the system and P be any point in the

In [14], the authors studied all linear maps Φ on B(H ) preserving in both direc- tions semi-Fredholm operators. It has been shown that such maps Φ preserve in both directions the

• Completion of a subset of all explicit tasks bound to a given parallel region may be specified through the use of task synchronization constructs. – Completion of all explicit

Let be a function of two variables for which the first- and second-order partial derivatives are continuous on some disk containing the point To apply the second derivative test

The purpose of this research is to identify if Roşia Montană mining project is according to the concept of sustainable development, by reaching the following Research

Let M be a ﬁve-dimensional proper nearly totally geodesic contact CR−submanifold of seven-dimensional unit sphere.. 5-dimensional contact CR submanifolds in

2.4 Vision localization of Truck picking point.. The having an innate capacity visible accompanying eyes place of residence or business where one can be contacted or special