Several inequalities of Erdős-Mordell type

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,R_a, R_b, R_c 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+R₃≥2 (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+z²R3≥2 (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+λ3R3≥2(√

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). 1≥_a^c ·_2R^R3_b +^b_a·_2R^R2_c b). 1≥_a^b ·_2R^R3_b +_a^c ·_2R^R2_c

Proof.

a). We use the well-known inequality

aR1≥cr2+br3

and the equality

r1= R2R3

2R_a who result writing the area of triangle MBC

S_△ABC =R2R3

4R_a = 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 =pr_AB^M , V =pr^M_AC. u=µ(∡U AM), v=µ(∡V AM).We have r₂+r₃=R₁(sinu+ sinv) = 2R₁sinA

2 cosu−v

2 ≤2R₁sinA 2 or

R3R1

2R_b +R1R2

2R_c ≤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, λ₃≥0.We have:

a). λ₁R^k_a+λ₂R^k_b+λ₃R^k_c ≥√

λ₂λ₃R^k₁+√

λ₃λ₁R₂^k+√ λ₁λ₂R^k₃ b). √

λ₂λ_3Rk^Rk1

b+R^k_c +√

λ₃λ_1Rk^Rk2

c+R_a^k+√

λ₁λ_2Rk^Rk3

a+R^k_b ≤

≤ ¹₂ (√

λ2λ3 R^k₁

√R^k_bR^k_c +√ λ3λ1

R^k₂

√R^k_cR^k_a +√ λ1λ2

R^k₃

√R^k_aR^k_b

)

≤ ^λ1+λ₂^2+λ3 Proof. a). From the inequality(x+y)^k≥2^k−1(

x^k+y^k)

, k∈[0,1]it result that:

(a c · R2

2R_a +b c · R1

2R_b )k

≥2^k−1 ([(a

c

)k R^k₂ (2Ra)^k +

(b c

)k( R1

2R_b )k])

Using this inequality and Lemma 1. We obtain

1≥ (a

c · R2

2R_a +b c· R1

2R_b )k

≥2^k−1 [(a

c

)k R^k₂ (2R_a)^k +

(b c

)k( R^k₁ 2R_b

)k]

or

2≥a^k c^k ·R₂^k

R_a^k +b^k c^k ·R^k₁

R^k_b or

2λ3R^k_c ≥λ3

a^k c^k · R^k_c

R^k_a ·R^k₂+λ3

b^k c^k ·R^k_c

R^k_b ·R^k₁ and the similar inequalities

2λ₂R^k_b ≥λ₂c^k b^k ·R^k_b

R^k_c ·R^k₁+λ₂a^k b^k ·R^k_b

R^k_a ·R^k₃

2λ₁R^k_a ≥λ₁b^k s^k ·R^k_a

R^k_b ·R^k₃+λ₁c^k a^k · R^k_a

R^k_c ·R^k₂ Adding this inequalities and apply the A-M inequalities we obtain:

2(

λ₁R^k_a+λ₂R_b^k+λ₃R^k_c)

≥R^k₁ (

λ₃·b^k c^k · R^k_c

R^k_b +λ₂· c^k b^k · R^k_b

R^k_c )

+

+R₂^k (

λ₃·a^k c^k ·R^k_c

R^k_a +λ₁· c^k a^k · R^k_a

R^k_c )

+R^k₃ (

λ₂· a^k b^k ·R^k_b

R^k_a +λ₁· b^k a^k ·R^k_a

R^k_b )

≥

≥2√

λ2λ3R₁^k+ 2√

λ3λ1R^k₂+ 2√ λ2λ1R^k₃

b). Adding the inequalities

2λ3≥λ3·a^k c^k · R^k₂

R^k_a +λ3· b^k c^k ·R^k₁

R^k_b

2λ1≥λ1· b^k a^k ·R^k₃

R^k_b +λ1· c^k a^k · R^k₂

R^k_c

2λ3≥λ2·c^k b^k ·R^k₁

Rc+λ2·a^k b^k ·R^k₃

R^k_a We obtain

2 (λ₁+λ₂+λ₃)≥ (

λ₃·a^k c^k · R^k₂

R^k_a +λ₁· c^k a^k ·R^k₂

R^k_c )

+

+ (

λ₃·b^k c^k ·R^k₁

R^k_b +λ₂·c^k b^k ·R^k₁

R^k_c )

+ (

λ₁· b^k a^k ·R^k₃

R^k_b +λ₂· a^k b^k ·R^k₃

R^k_a )

≥

≥2√

λ1λ3· R^k₂

√R^k_aR^k_c + 2√

λ2λ3· R^k₁

√ R^k_bR^k_c

+ 2√

λ1λ2· R^k₃

√ R^k_aR^k_b

≥

≥4√

λ₁λ₃· R^k₂

R^k_a+R^k_c + 4√

λ₂λ₃· R^k₁

R^k_b+R^k_c + 4√

λ₁λ₂· R^k₃ R_a^k+R^k_b

We observe that b) ^imply⇒ a)

Indeed putting in b). λ1→λ1R^k_a, λ2→λ2R^k_b, λ3→λ3R^k_c,we obtain

√

R^k_bR_c^k· R^k₁·√ λ2λ3

R_b^k+R_c^k +

√

R^k_cR^k_a· R^k₂·√ λ3λ1

R^k_a+R^k_c +

√

R^k_aR^k_b ·R^k₃·√ λ1λ2

√

R^k_a+R^k_b

≤

1 2

(λ₁R^k₁+λ₂R^k₂+λ₃R^k₃)

≤1 2

(λ₁R_a^k+λ₂R^k_b +λ₃R_c^k)

Also putting in b). λ₁→λ₁R^k_b, λ₂→λ₂R^k_c, λ₃→λ₃R^k_a andλ₁→λ₂R^k_b, λ₂→λ₃R^k_c, λ₃→λ₁R^k_a,we obtain:

Theorem 2. Let beM an interior point of triangleABC. Then for ecery real numberk∈[0,1], λ₁, λ₂, λ₃≥0, we have:

a).

√λ₂λ₃·R^k₁√ R^k_c ·R^k_a R^k_b+R^k_c +√

λ₃λ₁· R^k₂

√ R^k_a·R^k_b R^k_a+R^k_c +√

λ₁λ₂·R^k₃

√ R^k_b ·R^k_c R^k_a+R^k_b ≤

≤1 2



√ λ2λ3R^k₁

√ R^k_a R^k_b +√

λ3λ1R^k₂

√ R^k_b R^k_c +√

λ1λ2R₃^k

√ R^k_c R^k_a



≤

≤ 1 2

(λ1R^k_b +λ2R^k_c+λ3R^k_a) b).

√λ3λ1·R₁^k√ R^k_c ·R^k_a R^k_b+R^k_c +√

λ1λ2· R^k₂

√ R^k_a·R_b^k R^k_c +R^k_a +√

λ2λ3·R^k₃

√ R_b^k·R^k_c R^k_a+R^k_b ≤

≤1 2



√ λ₃λ₁R^k₁

√ R^k_a R^k_b +√

λ₁λ₂R^k₂

√ R^k_b R^k_c +√

λ₂λ₃R^k₃

√ R^k_c R^k_a



≤

≤ 1 2

(λ₂R^k_b +λ₃R^k_c+λ₁R^k_a)

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

Corollary 2.1. a).

R^k_a+R^k_b +R^k_c ≥R^k₁+R^k₂+R^k₃ b).

R^k₁

R^k_b +R^k_c + R^k₂

R^k_c +R^k_a + R^k₃

R^k_a+R^k_b ≤ 1 2



 R^k₁

√ R^k_bR^k_c

+ R^k₂

√R^k_cR^k_a + R₃^k

√ R^k_aR^k_b



≤3 2

c).

R^k₁√ R^k_cR^k_a R^k_b +R_a^k +

R^k₂

√ R^k_aR_b^k R^k_a+R^k_c +

R^k₃

√ R^k_bR^k_c R^k_a+R^k_b ≤ 1

2



R^k₁

√ R^k_a R^k_b +R^k₂

√ R^k_b R^k_c +R^k₃

√ R^k_c R^k_a



≤

≤ 1 2

(R^k_a+R^k_b +R^k_c)

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

R_b+R_c + R2

R_c+R_a + R3

R_a+R_b ≤1 2

( R1

√RbRc

+ R2

√RaRb

+ R3

√RaRb

)

who represent a refinement of inequality _R^R1

b+R_c +_R^R2

c+R_a +_R^R3

a+R_b ≤³₂ who apperas in [2].

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

√RcRa

R_b+R_c +R2

√RaRb

R_a+R_c +R3

√RbRc

R_a+R_c ≤ 1 2

( R1

√Ra

R_b +R2

√Rb

R_c +R3

√Rc

R_a )

≤

≤ 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, λ3≥0, we have

λ²₁R^k_a+λ²₂R^k_b +λ²₃R²_c ≥λ1λ2λ3

(R^k₁ λ₁ +R^k₂

λ₂ +R^k₃ λ₃

) If we take k= 1we obtain

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).

R^k₁R^k_a+R^k₂R^k_b +R^k₃R^k_c ≤R^k_aR_b^k+R^k_bR^k_c +R_c^kR^k_a b).

1 + R^k_b R^k_c +R^k_c

R^k_b ≥ R^k₁

√ R^k_bR^k_c

+ R^k₂

√ R^k_aR^k_b

+ R^k₃

√R^k_aR^k_c

and similar inequalities c).

R^k₁R^k_a+R₂^k

√

R^k_bR^k_c +R^k₃

√

R^k_bR^k_c ≤R^k_aR^k_b +R^k_bR^k_c +R^k_cR^k_b and similar inequalities

d).

R^k_a R^k_b +R^k_b

R^k_c +R_c^k

R_a^k ≥ R^k₁

√R^k_cR^k_a + R^k₂

√ R^k_bR^k_a

+ R^k₃

√ R^k_bR_c^k

and similar inequalities e).

R^k₁

√

R_a^kR^k_b +R^k₂

√

R^k_bR^k_c +R^k₃

√

R^k_aR^k_c ≤R^k_aR^k_b +R^k_bR^k_c +R^k_cR^k_b Proof. We take in theorem 1

(λ₁, λ₂, λ₃) = ( 1

R_a^k, 1 R_b^k, 1

R_c^k )

,(λ₁, λ₂, λ₃) = ( 1

R^k_a, 1 R^k_b, 1

R^k_c )

(λ1, λ2, λ3) = ( 1

R_b^k, 1 R_a^k, 1

R_c^k )

,(λ1, λ2, λ3) = ( 1

R^k_b, 1 R^k_c, 1

R^k_a )

(λ₁, λ₂, λ₃) = ( 1

R_c^k, 1 R_a^k, 1

R_b^k )

,(λ₁, λ₂, λ₃) = ( 1

R^k_c, 1 R^k_b, 1

R^k_a )

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

a).

R^k₁ R^k_b +R₂^k

R_c^k +R^k₃ R^k_a ≤R_b^k

R_c^k +R^k_c R^k_a +R^k_a

R^k_b

b).

R^k₁ R^k_b

√ R^k_a R^k_c +R^k₂

R^k_c

√ R^k_b R^k_a +R^k₃

R^k_a

√ R^k_c R^k_b ≤R_b^k

R_a^k +R^k_c R^k_b +R^k_a

R^k_c

c).

R^k₁ R^k_b +R₂^k

R_c^k

√ R^k_b

R^k_a + R^k₃

√ R^k_bR^k_a

≤R^k_b R^k_c +R^k_c

R^k_b + 1

and similar inequalities d).

R^k₁

√ R^k_bR^k_c

+ R^k₂

√R^k_cR^k_a + R^k₃

√ R^k_aR^k_b

≤3

Proof. We take in Theorem 2:

(λ₁, λ₂, λ₃) = ( 1

R_a^k, 1 R^k_b, 1

R^k_c )

,(λ₁, λ₂, λ₃) = ( 1

R^k_a, 1 R^k_b, 1

R^k_c )

(λ1, λ2, λ3) = ( 1

R_b^k, 1 R^k_a, 1

R^k_c )

,(λ1, λ2, λ3) = ( 1

R^k_b, 1 R^k_c, 1

R^k_a )

(λ1, λ2, λ3) = ( 1

R_c^k, 1 R^k_a, 1

R^k_b )

,(λ1, λ2, λ3) = ( 1

R^k_c, 1 R^k_b, 1

R^k_a )

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

a).

R₁^2kR^k_c +R^2k₂ R^k_a+R^2k₃ R^k_b ≥R^2k₁ R₃^k+R^2k₂ R^k₁+R₃^2kR^k₂ b).

R^2k₁ R^k₃

√ R^k_bR_c^k

+ R^2k₂ R^k₁

√R^k_cR^k_a + R^2k₃ R^k₂

√ R_a^kR^k_b

≤R^2k₁ +R^2k₂ +R^2k₃

c).

R₁^2kR^k₃

√ R^k_a

R^k_b +R^2k₂ R₁^k

√ R^k_b

R^k_c +R^2k₃ R^k₂

√ R^k_c

R^k_a ≤R^2k₂ R_b^k+R^2k₃ R^k_c +R₁^2kR^k_a d).

R^2k₁ R^k₂

√ R^k_a

R^k_b +R^2k₂ R^k₃

√ R^k_b

R^k_c +R^2k₃ R₁

√ R^k_c

R^k_a ≤R^2k₃ R^k_b +R^2k₁ R^k_c +R^2k₂ R^k_a Proof. We take

(λ1, λ2, λ3) = ( R^k₂

R^k₁R^k₃, R^k₃ R^k₁R^k₂, R₁^k

R^k₂R^k₃, )

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).

R^k_a R^2k₂ + R^k_b

R^2k₃ + R^k_c R^2k₁ ≥ 1

R₁^k + 1 R^k₂ + 1

R^k₃ b).

1

R^k₃

√ R^k_bR^k_c

+ 1

R^k₁√

R^k_cR^k_a + 1 R₂^k

√ R^k_aR^k_b

≤ 1 R^2k₁ + 1

R^2k₂ + 1 R^2k₃

Proof. We take in Theorem 1:

(λ1, λ2, λ3) = ( 1

R^2k₂ , 1 R^2k₃ , 1

R^2k₁ )

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

RaRbRc

R1R2R3 ≥ 1 8^1k ·

∏ (b^k+c^k)1/k

abc ≥1 b).

∏ (R^k₃ R^k_b +R^k₂

R^k_c )

≤ 64 (abc)^k

∏(b^k+c^k) ≤8

Proof. From the inequalitiesR1≥ ^c_ar2+_a^br3, R1≥ _a^br2+^c_ar3 and(x+y)^k≥2^k−1(

x^k+y^k) ,if k∈[0,1]it result.

R^k₁≥ (b

ar2+ c ar3

)k

≥2^k−1 (b^k

a^kr^k₂+c^k a^kr₃^k

)

and

R^k₁≥ (c

ar2+ b ar3

)k

≥2^k−1 (c^k

a^kr^k₂+b^k a^kr₃^k

)

Adding the last two inequalities we obtain

2R^k₁ ≥2^k−1 [

r^k₂

(b^k+c^k a^k

) +r^k₃

(b^k+c^k a^k

)]

= 2^k−1(

r₂^k+r^k₃) (

b^k+c^k) a^k

and the similar inequalities.

By multiplication we obtain:

8 (R1R2R3)^k≥8^k−1

∏ (r₂^k+r^k₃) (

b^k+c^k) a^kb^kc^k Taking account by equalitiesr2=^R_2R^3R1

b , r3= ^R_2R^1R2

c we obtain:

8 (R1R2R3)^k≥8^k−1∏ (R^k₃R₁^k

2^kR_b^k +R^k₁R^k₂ 2^kR^k_c

) ∏ (

b^k+c^k) a^kb^kc^k or

64≥∏ (R^k₃ R^k_b +R₂^k

R_c^k ) ∏ (

b^k+c^k)

a^kb^kc^k ≥8∏ (R^k₃ R^k_b +R^k₂

R^k_c )

or

∏ (R^k₃ R^k_b +R^k₂

R^k_c )

≤ 64a^kb^kc^k

∏(b^k+c^k) ≤8

b). We have

64≥∏ (R^k₃ R^k_b +R^k₂

R^k_c ) ∏ (

b^k+c^k) a^kb^kc^k ≥8

∏ (b^k+c^k)

a^kb^kc^k · (R1R2R3)^k (RaRbRc)^k.