# Several inequalities of Erdős-Mordell type

## Full text

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### 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

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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 )

+

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+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

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).

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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

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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).

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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).

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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.

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