INTERNATIONAL JOURNAL OF GEOMETRYVol. 9 (2020), No. 2, 81-92

Two new refinements of a linear geometric inequality

Jian Liu

Abstract. For a consequence of the Erdos-Mordell inequality, we estab-lish two new refinements with the help of Maple software. Some applicationsare also given by new results. Several interesting conjectures checked bycomputer are presented as well.

1. Introduction

The famous Erdos-Mordell inequality is a beautiful linear geometric in-equality, which can be stated as follows:

Let P be an interior point of the triangle ABC. Denote by R1, R2, R3 thedistances from P to the vertices A,B,C respectively, and denote by r1, r2, r3the distances from P to the sides BC,CA,AB ,respectively. Then

(1) R1 +R2 +R3 ≥ 2(r1 + r2 + r3),

with equality if and only if △ABC is equilateral and P is its center.There exist various proofs, generalizations, sharpness and refinements of

the Erdos-Mordell inequality in the literature. Some recent results on theErdos-Mordell inequality can be found in [2],[6]-[10].

By the Erdos-Mordell inequality, it is easy to obtain the following inequal-ity:

(2) R1 +R2 +R3 ≥2

3(ha + hb + hc),

where ha, hb, hc are the altitudes of the triangle ABC. In fact, by the triangleinequality we have

(3) R1 + r1 ≥ ha

and two similar relations. Adding 2(R1 +R2 +R3) to both sides of (1) andthen using inequality (3) etc., we obtain one inequality equivalent with (2)immediately.

Clearly, inequality (2) can be regarded as a consequence of the Erdos-Mordell inequality. We wish to point out that it is valid for any point P inthe plane of the triangle ABC, because we have the following conclusion:————————————–

Keywords and phrases: the Erdos-Mordell inequality, Triangle, Point,Altitude, Median

(2010)Mathematics Subject Classification: 51M16Received: 29.04.2020. In revised form: 10.07.2020. Accepted: 07.06.2020.

82 Jian Liu

For any pointQ outside triangle ABC, there exist a point P inside triangleABC such that QA ≥ PA,QB ≥ PB,QC ≥ PC at the same time. Thisconclusion is not difficult to be proved, we omit here.

From [12], we know that Ji Chen conjectured that the following inequalitystronger than (2) holds:

(4) R1 +R2 +R3 ≥2

3(wa + wb + wc),

where wa, wb, wc are the lengths of angle bisectors of triangle ABC.In [16], Xiao-Guang Chu proved inequality (4). As wa ≥ ha etc., thus we

have the following refinement of inequality (2):

(5) R1 +R2 +R3 ≥2

3(wa + wb + wc) ≥

2

3(ha + hb + hc).

In [4], the author and Xiao-Guang Chu established the following inequal-ity:

R1 +R2 +R3 ≥1

2(ma +mb +mc + 3r),(6)

where ma,mb,mc are the lengths of medians of △ABC and r is the inradiusof ABC.

By inequality (6), we can further obtain the following refinement of in-equality (2):

R1 +R2 +R3 ≥1

2(ma +mb +mc + 3r)

≥ 1

3(ma +mb +mc + ha + hb + hc) ≥

2

3(ha + hb + hc).(7)

where the second inequality has been proved in [4] and the last inequalityis obvious by ma ≥ ha etc.

Our aim of this paper is to give two new refinements of the linear geometricinequality (2).

In what follows, we shall continue to use the previous symbols. Also,denote by a, b, c the side lengths of the triangle ABC and denote by s,R, rthe sem-perimeter, the circumradius and the inradius of the triangle ABCrespectively.

The main results are as follows:

Theorem 1.1. For any point P in the plane of the triangle ABC, it holds:

(8) R1 +R2 +R3 ≥√3

2a+ ha ≥ 2

3(ha + hb + hc),

where the first equality in (8) holds if and only if b = c, P lies on the altitude

from the vertex A to the side BC and ∠BPC =2

3π, the second inequality

holds if and only if the triangle ABC is equilateral.

Theorem 1.2. For any point P in the plane of the triangle ABC, it holds:

R1 +R2 +R3 ≥ 2r +2

3

√a2 + b2 + c2 ≥ 2

3

(1 +

r

R

)√a2 + b2 + c2

≥ 2

3(ha + hb + hc),(9)

Two new refinements of a linear geometric inequality 83

where the first equality in (9) holds if and only if the triangle ABC is equi-lateral and P is its center, the others hold if and only if the triangle ABCis equilateral.

2. Proof of Theorem 1.1

Before proving Theorem 1.1, we first give the following lemma.

Lemma 2.1. In any triangle ABC, holds:

(10) b+ c ≥√3

2a+ ha,

with equality if and only if b = c and A =2

3π.

Proof. By the known formula

(11) ha =

√s(s− a)(s− b)(s− c)

a

(where s = (a+ b+ c)/2) and the fact that√

(s− b)(s− c) ≤ 2a, we have

(12) ha ≤√s(s− a).

Squaring both sides and using s = (a+ b+ c)/2, we easily get

(13) b+ c ≥√

a2 + 4h2a.

But

(14) a2 + 4h2a =

(√3

2a+ ha

)2

+

(1

2a−

√3ha

)2

,

thus from (13) and (14) we see that (10) holds. Observe that the equalityin (13) holds if and only if b = c, thus by (13) and (14) one sees that theequality in (10) holds if and only if b = c and a/2−

√3ha = 0, from which

and aha = 2S (S being the area of ABC) it follows that a4 = 12S2. Further,by the equivalent form of Heron’s formula:

(15) 2b2c2 + 2c2a2 + 2a2b2 − a4 − b4 − c4 = 16S2,

we get

2b2c2 + 2c2a2 + 2a2b2 − a4 − b4 − c4 =4

3a4.

Substituting b = c into the above identity, we easily get a =√3c and then

conclude that B = C = π/3 and A = 2π/3. Hence the equality conditionof (10) is just as mentioned as in Lemma 2.1. This completes the proof ofLemma 2.1.

Next, we prove Theorem 1.1.Proof. Firstly, we prove the first inequality of (9), i.e.,

(16) R1 +R2 +R3 ≥√3

2a+ ha.

Applying inequality (10) to triangle BPC, we have

(17) R2 +R3 ≥√3

2a+ r1.

84 Jian Liu

Adding R1 to both sides of this inequality and then using inequality (3), oneobtains inequality (16) immediately. Note that the equality in (3) holds ifand only if P lies on the altitude from the vertex A to the side BC. Thus bythe equality condition of (10), we further know that the equality conditionof (16) is just as mentioned in Theorem 1.1.

Next, we prove the second inequality of (8), i.e.,

(18)

√3

2a+ ha ≥ 2

3(ha + hb + hc).

This inequality is equivalent to

(19)3√3

2a+ ha ≥ 2(hb + hc).

Taking squares on both sides and then using aha = 2S = 2sr, it becomes

(20)27

4a2 + h2a + 6

√3rs ≥ 4(hb + hc)

2.

Multiplying both sides by 4 and noticing the following known inequality

(21) s ≥ 3√3r,

we only need to prove that

(22) 27a2 + 4h2a + 216r2 ≥ 16(hb + hc)2,

i.e.,

27a2 +16S2

a2+

216S2

s2≥ 64S2

(1

b+

1

c

)2

.

And, by Heron formula:

(23) S =√

s(s− a)(s− b)(s− c),

we easily know again the inequality is equivalent to

Q0 ≡ 4(s− a)(s− b)(s− c)[16b2c2s2 + 216(abc)2 − 64s2a2(b+ c)2

]+ 108sa4b2c2 ≥ 0.(24)

Substituting s = (a+ b+ c)/2 into (24) and then arranging with the help ofMaple software, we know again that (24) is equivalent to

Q0 ≡4(b+ c)2a7 + 4(b+ c)3a6 − (8b4 + 16b3c+ 44b2c2 + 16bc3 + 8c4)a5

− 8(b+ c)(b4 + 2b3c− 8b2c2 + 2bc3 + c4)a4 + (4b6 + 8b5c+ 52b4c2

− 124b3c3 + 52b2c4 + 8bc5 + 4c6)a3 + 4(b+ c)(b6 + 2b5c− 14b4c2

+ 23b3c3 − 14b2c4 + 2bc5 + c6)a2 − (b− c)2(b+ c)2b2c2a

− (b− c)2(b+ c)3b2c2 ≥ 0,(25)

which is required to prove.Now, we let b+c−a = 2x, c+a−b = 2y, a+b−c = 2z, then a = y+z, b =

z + x, c = x+ y. Substituting them into Q0 and using Maple software, it iseasy to get the following identity:

(26) Q0 = 2Q1,

Two new refinements of a linear geometric inequality 85

where

Q1 ≡ 16yzx7 + 64(y + z)yzx6 + (27y4 + 164y3z + 306y2z2 + 164yz3

+ 27z4)x5 + (y + z)(81y4 + 52y3z + 38y2z2 + 52yz3 + 81z4)x4

+ (81y6 − 60y5z − 441y4z2 − 584y3z3 − 441y2z4 − 60yz5

+ 81z6)x3 + (y + z)(27y6 − 114y5z − 235y4z2 − 156y3z3

− 235y2z4 − 114yz5 + 27z6)x2 − yz(10y4 + 13y3z − 226y2z2

+ 13yz3 + 10z4)(y + z)2x+ 27(y + z)5y2z2.

Thus, it remains to prove that Q1 ≥ 0 holds for positive real numbers x, y, z.Since Q1 is symmetric in y and z, we may assume without loss of generalitythat y ≥ z and let y = z +m(m ≥ 0). Substituting y = z +m into Q1 andusing Maple software, we immediately obtain the following identity:

Q1 = (27z2 − 10zx+ 27x2)m7 + (324z3 − 103z2x+ 102zx2 + 81x3)m6

+ (1647z4 − 218z3x− 304z2x2 + 426zx3 + 81x4)m5

+ (4590z5 + 531z4x− 2496z3x2 + 474z2x3 + 538zx4 + 27x5)m4

+ 8z(945z5 + 348z4x− 780z3x2 − 166z2x3 + 179zx4 + 34x5)m3

+ 16z(459z6 + 268z5x− 506z4x2 − 264z3x3 + 123z2x4

+ 60zx5 + 4x6)m2 + 16z(z − x)(z + x)(243z5 + 180z4x

− 107z3x2 − 87z2x3 − 12zx4 − x5)m+ 16(54z3 + 45z2x

+ 8zx2 + x3)(z − x)2(z + x)2z2.(27)

To show that inequality Q1 ≥ 0 is valid for y > 0, z > 0 and m ≥ 0, we shallconsider two cases to finish the proof.

Case 1. z ≥ x.In this case, we set z = x+n(n ≥ 0) and substitute it into (27), then the

inequality Q1 ≥ 0 becomes

Q2 ≡ (2304m2 + 6912mn+ 6912n2)x7 + 128(m+ 2n)(35m2 + 146mn

+ 146n2)x6 + (3664m4 + 33472m3n+ 119104m2n2 + 171264mn3

+ 85632n4)x5 + 32(m+ 2n)(51m4 + 494m3n+ 2176m2n2

+ 3364mn3 + 1682n4)x4 + (404m6 + 5752m5n+ 42072m4n2

+ 152752m3n3 + 276656m2n4 + 240336mn5 + 80112n6)x3

+ 4(m+ 2n)(11m6 + 195m5n+ 1841m4n2 + 7700m3n3

+ 14870m2n4 + 13224mn5 + 4408n6)x2 + n(m+ n)(44m4

+ 649m3n+ 2773m2n2 + 4248mn3 + 2124n4)(m+ 2n)2x

+ 27(m+ n)2(m+ 2n)5n2 ≥ 0.(28)

Since m ≥ 0, n ≥ 0 and x > 0, we see that Q2 ≥ 0 holds.Case 2. x > z.

86 Jian Liu

In this case, we set x = z+n(n ≥ 0) and substitute it into (27), then theinequality Q1 ≥ 0 becomes

Q3 ≡ (2304m2 − 6912mn+ 6912n2)z7 + (4480m3 − 11520m2n

+ 7680mn2 + 11008n3)z6 + (3664m4 − 6592m3n+ 1600m2n2

+ 19968mn3 + 6528n4)z5 + 16(3m+ 4n)(34m4 − 61m3n+ 104m2n2

+ 164mn3 + 28n4)z4 + (404m6 + 776m5n+ 2424m4n2 + 7120m3n3

+ 7728m2n4 + 2864mn5 + 240n6)z3 + 4(m+ n)(11m6 + 75m5n

+ 290m4n2 + 434m3n3 + 264m2n4 + 72mn5 + 4n6)z2

+mn(44m4 + 257m3n+ 192m2n2 + 32mn3 + 16n4)(m+ n)2z

+ 27(m+ n)3m4n2 ≥ 0.(29)

Since

2304m2 − 6912mn+ 6912n2 > 0,

4480m3 − 11520m2n+ 7680mn2 = 640m(7m2 − 18mn+ 12n2) > 0,

3664m4 − 6592m3n+ 1600m2n2 = 16m2(229m2 − 412mn+ 100n2) > 0,

34m4 − 61m3n+ 104m2n2 = m2(34m2 − 61mn+ 104n2) > 0,

we conclude that Q3 ≥ 0 holds for m ≥ 0, n ≥ 0, and z > 0 from (29).Combining the arguments of the above two cases, inequality Q1 ≥ 0 is

proved for all positive numbers x, z and non-positive real number m. Hence,inequality Q0 ≥ 0 and inequality (18) are proved too. Also, it is easy toconclude that the equality in (18) holds if and only if a = b = c, i.e., △ABCis equilateral. This completes the proof of Theorem 1.1.

Remark 2.1. By inequality (17), we have

(30)R2 +R3 − r1

a+

R3 +R1 − r2b

+R1 +R2 − r3

c≥ 3

√3

2

which is a special case of a conjecture posed by the author in [5].

Remark 2.2. By inequality (17) and Gerber’s inequality (see [3]):

(31) r2r3 + r3r1 + r1r2 ≤S√3,

we easily obtain the following known inequality (see [11]):

R1(r2 + r3) +R2(r3 + r1) +R3(r1 + r2)

≥ (r3 + r1)(r1 + r2) + (r1 + r2)(r2 + r3) + (r2 + r3)(r3 + r1).(32)

3. Proof of Theorem 1.2

In order to prove Theorem 1.2, we shall use the following known result(see [1, inequality 12.55]):

Lemma 3.1. Let P be a point in the plane of the triangle ABC.

(i) If A ≥ 2

3π, then

(33) R1 +R2 +R3 ≥ b+ c,

Two new refinements of a linear geometric inequality 87

with equality if and only if P = A.

(ii) If A <2

3π, then

(34) R1 +R2 +R3 ≥√

1

2(a2 + b2 + c2) + 2

√3S,

with equality if and only if ∠BPC = ∠CPA = ∠APB.

Lemma 3.2. In triangle ABC, if max{A,B,C} ≤ 2

3π then

(35)√3s ≥ 2R+ 5r,

with equality if and only if △ABC is equilateral.

Inequality (35) can be obtained by Xue-Zhi Yang’s result in [15], whichwas used to prove inequality (4) by Xiao Chu in [16].

We are now read to prove Theorem 1.2.Proof. We first prove the first inequality of (9), i.e.,

(36) R1 +R2 +R3 ≥ 2r +2

3

√a2 + b2 + c2.

Based on Lemma 1, we divide the following two cases to complete the proof.

Case 1. A >2

3π.

In this case, by Lemma 1 we only need to prove that

(37) b+ c > 2r +2

3

√a2 + b2 + c2.

Note that b+ c > 2ha > 4r, thus we only need to show that

9(b+ c− 2r)2 − 4(a2 + b2 + c2) > 0,

i.e.,

(38) 9(b+ c)2 + 36r2 − 4a2 − 4b2 − 4c2 > 36r(b+ c).

Multiplying both sides by s and using S = rs and the known identity

(39) (s− a)(s− b)(s− c) = sr2,

we know that inequality (38) is equivalent to

s(5b2 + 5c2 + 18bc− 4a2

)+ 36(s− a)(s− b)(s− c)

> 36(b+ c)√s(s− a)(s− b)(s− c).(40)

For proving the above inequality, we set s− a = x, s− b = y, s− c = z, thena = y + z, b = z + x, c = x + y, s = x + y + z(x, y, z > 0) and we see thatinequality (40) is equivalent to

(x+ y + z)[5(z + x)2 + 5(x+ y)2 + 18(z + x)(x+ y)− 4(y + z)2

]+ 36xyz > 36(2x+ y + z)

√xyz(x+ y + z),

i.e.,

28x3 + 56(y + z)x2 + (29y2 + 102yz + 29z2)x+ (y + z)(y2 + 10yz + z2)

> 36(2x+ y + z)√

xyz(x+ y + z).

88 Jian Liu

Thus, we need to prove[28x3 + 56(y + z)x2 + (29y2 + 102yz + 29z2)x+ (y + z)(y2 + 10yz + z2)

]2− 362xyz(x+ y + z)(2x+ y + z)2 > 0.

Expanding out and collecting like terms gives

784x6 + 3136(y + z)x5 + (4760y2 + 6800yz + 4760z2)x4

+8(y + z)(413z2 + 202yz + 413y2)x3 + (953y4 + 780y3z

+1590y2z2 + 780yz3 + 953z4)x2 + 2(y + z)(29y4 − 256yz3

−218y2z2 − 256y3z + 29z4)x+ (y + z)2(y2 + 10yz + z2)2 > 0.(41)

Obviously, to prove this inequality it is enough to prove that

(953y4 + 780y3z + 1590y2z2 + 780yz3 + 953z4)x2

+ 2(y + z)(29y4 − 256yz3 − 218y2z2 − 256y3z + 29z4)x

+ (y + z)2(y2 + 10yz + z2)2 > 0.(42)

The left is a quadratic function on x and its discriminant Fx can easily becalculated as follows:

Fx = −64(7y6+2154y5z−471y4z2+940y3z3−471y2z4+2154yz5+7z6)(y+z)4.

Since

2154y5z − 471y4z2 − 471y2z4 + 2154yz5

= 3yz[(561(y4 + z4) + 157(y2 + yz + z2)(y − z)2

]> 0,

so that Fx < 0, thus we conclude that inequality (42) holds and inequality(37) is proved.

Case 2 A ≤ 2

3π.

In this case, by Lemma 2, to prove inequality (36) we only need to provethat

(43)

√1

2(a2 + b2 + c2) + 2

√3S ≥ 2r +

2

3

√a2 + b2 + c2.

Squaring both sides, it becomes

1

2(a2 + b2 + c2) + 2

√3S ≥ 4r2 +

4

9(a2 + b2 + c2) +

8

3r√

a2 + b2 + c2,

that is(a2 + b2 + c2) + 36

√3rs ≥ 72r2 + 48r

√a2 + b2 + c2.

Squaring both sides again, we have to prove that

(a2 + b2 + c2)2 + 3888r2s2 + 72√3rs(a2 + b2 + c2)

≥ 5184r4 + 2304r2(a2 + b2 + c2) + 6912r3√

a2 + b2 + c2.(44)

According to Lemma 3 and the following known identity:

(45) a2 + b2 + c2 = 2(s2 − 4Rr − r2),

we only need to prove the following inequality:

4(s2 − 4Rr − r2)2 + 3888r2s2 + 144(2R+ 5r)r(s2 − 4Rr − r2)

≥ 5184r4 + 4608r2(s2 − 4Rr − r2) + 6912r3√

2(s2 − 4Rr − r2),

Two new refinements of a linear geometric inequality 89

i.e.,

4s4 + (256Rr − 8r2)s2 − 1088R2r2 + 15296Rr3 − 1292r4

≥ 6912r3√

2(s2 − 4Rr − r2).(46)

According to the Gerretsen inequality (see [1, inequality 5.8])

(47) s2 ≥ 16Rr − 5r2

and Euler inequality R ≥ 2r, we have

4s4 + (256Rr − 8r2)s2 − 1088R2r2 + 15296Rr3 − 1292r4

> 4(16Rr − 5r2)2 + (256Rr − 8r2)(16Rr − 5r2)− 1088R2r2

+ 15296Rr3 − 1292r4

= 576r2(7R2 + 23Rr − 2r2) > 0.

Thus, to prove (46) we only need to prove that[4s4 + (256Rr − 8r2)s2 − 1088R2r2 + 15296Rr3 − 1292r4

]2−[6912r3

√2(s2 − 4Rr − r2)

]2≥ 0.

After arranging, it becomes

Q0 ≡ 16s8 + 64r(32R− r)s6 + 96(592R2 + 1232Rr − 107r2)r2s4

− 64(8704R3 − 122640R2r + 14160Rr2 + 1492669r3)r3s2

+ 16(73984R4 − 2080256R3r + 14798688R2r2

+ 21417568Rr3 + 6076297r4)r4 ≥ 0,(48)

which is required to prove.Putting

e = R− 2r,

g1 = s2 − 16Rr + 5r2,

g2 = 4R2 + 4Rr + 3r2 − s2,

then we have e ≥ 0 and Gerretsen inequalities g1 ≥ 0, g2 ≥ 0(see [1]).Through analysis, we obtain the following identity:

(49) Q0 = m1g2 + e(m2g1 +m3) +m4g21,

where

m1 = 35831808r6,

m2 = 55296(56R2 + 289Rr + 539r2)r3,

m3 = 331776(49R3 + 420R2r + 909Rr2 − 434r3)r4,

m4 = 16[s4 + 2r(80R− 7r)s2 + (8416R2 + 5504Rr − 527r2)r2

].

Therefore, by Euler inequality R ≥ 2r and Gerretsen inequalities g1 ≥ 0,g2 ≥ 0, we see that Q0 ≥ 0 holds true.

Combining the arguments of the above two cases, we finish the proof ofinequality (36).

The last two inequalities of (9) are easily proved.

90 Jian Liu

The second inequality of (9) is actually equivalent to

(50)√

a2 + b2 + c2 ≤ 3R,

which is also equivalent to the simple known inequality a2 + b2 + c2 ≤ 9R2.Since bc = 2Rha, we see that the third inequality of (9) is equivalent to

(51) (R+ r)√

a2 + b2 + c2 ≥ bc+ ca+ ab.

In view of a2 + b2 + c2 ≥ bc + ca + ab and the following known inequality(see [1, inequality 5.17]):

(52) bc+ ca+ ab ≤ 4(R+ r)2,

we conclude that inequality (51) holds and finish the proof of inequalitychain (9). In addition, we can easily determine the equality conditions of(9). This completes the proof of Theorem 1.2.

4. Corollaries and conjectures

In this section, we give several corollaries of Theorem 1.1 and Theorem1.2 and present a few relevant conjectures.

Since (√3

2a+ ha

)2

≥ 2√3aha = 4

√3S,

thus by the first inequality of (8) we get the following known inequality (see[1, inequality 12.18]):

Corrolary 4.1. For any point P in the plane of △ABC, it holds:

(53) R1 +R2 +R3 ≥ 2

√√3S.

Let P be the centroid of △ABC, then R1 =2

3ma, R2 =

2

3mb, R3 =

2

3mc.

Thus by the first inequality of (8) we have

Corrolary 4.2. In any triangle ABC, holds:

(54) ma +mb +mc ≥3√3

4a+

3

2ha.

For the acute triangle ABC, the author conjectures that the followingstronger inequality holds.

Conjecture 4.1. In the acute triangle ABC, holds:

(55) ma +mb +mc ≥3√3

4a+

3

2wa.

For the first inequality of (8),i.e.,(10), we also have two similar inequali-ties. Adding these three inequalities gives

Corrolary 4.3. For any point P in the plane of △ABC, it holds:

(56) R1 +R2 +R3 ≥1

3(√3s+ ha + hb + hc).

Let P be the centroid of △ABC, then by (56) we can obtain

Two new refinements of a linear geometric inequality 91

Corrolary 4.4. In any triangle ABC, holds:

(57) 2(ma +mb +mc) ≥√3s+ ha + hb + hc.

It follows from inequalities (19) and (17) that

(58) 3(R2 +R3) ≥ 2(hb + hc)− ha + 3r1.

Adding R1 and then using the previous inequality (3), the following linearinequality is obtained:

Corrolary 4.5. For any point P in the plane of △ABC, it holds:

(59) R1 + 3(R2 +R3) ≥ 2(r1 + hb + hc),

Remark 4.1. The previous inequalities (3) and (16) are both actually validfor any point P in the plane, thus inequality (59) holds in the same situationtoo.

For inequality (59), we present the following stronger inequality.

Conjecture 4.2. For any point P in the plane of △ABC, it holds:

(60) R1 + 3(R2 +R3) ≥ 2(r1 + wb + wc).

Let P be the centroid of △ABC, then by the first inequality of (9) we get

Corrolary 4.6. In any triangle ABC, holds:

(61) ma +mb +mc ≥ 3r +√

a2 + b2 + c2.

Inspired by this inequality, we pose the following conjecture:

Conjecture 4.3. In any triangle ABC, holds:

(62) 3r +√

a2 + b2 + c2 ≥ ka + kb + kc,

where ka, kb, kc are the symmedians of △ABC.

Remark 4.2. If (62) is true then by Theorem 2 we conclude that the fol-lowing linear inequality similar to (2) and (4) holds:

(63) R1 +R2 +R3 ≥2

3(ka + kb + kc),

which was proved by Xiao-Guang Chu in [16].

Remark 4.3. For the acute triangle ABC, the author has already provedthe following inequality:

(64) 3r +√a2 + b2 + c2 ≥ wa + wb + wc,

which shows that the first inequality of (9) is stronger than inequality (4) forthe acute triangle ABC.

Adding inequality (17) and its two analogues gives

(65) 2(R1 +R2 +R3)− (r1 + r2 + r3) ≥√3s,

which is valid for any interior point P of triangle ABC. Here, we put forwardthe following similar conjecture:

Conjecture 4.4. For any interior point P of △ABC, it holds:

(66) 2(R1 +R2 +R3)− (r1 + r2 + r3) ≥ ma +mb +mc.

92 Jian Liu

Remark 4.4. If the above inequality holds, then by inequality (3) one easilyobtains

(67) R1 +R2 +R3 ≥1

3(ma +mb +mc + ha + hb + hc),

which follows from the previous inequality chain (7).

Finally, we present an inequality which is stronger than the Erdos-Mordellinequality (1).

Conjecture 4.5. For any interior point P of △ABC, it holds:

(68) R1 +R2 +R3 ≥ 2

(√ma

har1 +

√mb

hbr2 +

√mc

hcr3

).

References

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