Seven geometric inequalites proposed by tran quang hung

Some geometric inequalitiesTran Quang Hung Problem 1.. With arbitrary triangle ABC inscribed O, incenter I and an arbitrary point M in small arc BC.. Given are4ABC orthorcenter H, circum

Some geometric inequalities

Tran Quang Hung Problem 1 With arbitrary triangle ABC inscribed (O), incenter I and an arbitrary point M in small arc BC Prove that MA + 2OI ≥ MB + MC ≥ MA − 2OI

Proof By the projection of vectors we have

MA2

= 2−−→

MO.−−→

MA ⇒ MA = 2−−→MO

−−→

MA MA similar

MB = 2−−→MO.−−→MB

MB, MC = 2

−−→

MO

−−→

MC MC From them we have

MB + MC − MA = 2−−→MO.(

−−→

MB

MB +

−−→

MC

MC −

−−→

MA

MA) (1)

By Cauchy-Swart inequality we have

−MO|

−−→

MB

MB +

−−→

MC

MC −

−−→

MA

MA| ≤−−→MO · (

−−→

MB

MB +

−−→

MC

MC −

−−→

MA

MA) ≤ MO|

−−→

MB

MB +

−−→

MC

MC −

−−→

MA

MA| (2)

We have MO = R, we will calculate |

−−→

MB

MB +

−−→

MC

MC −

−−→

MA

MA|

|

−−→

MB

MB +

−−→

MC

MC −

−−→

MA

MA|2

= 3 + 2(

−−→

MB

MB.

−−→

MC

MC −

−−→

MB

MB.

−−→

MA

MA −

−−→

MC

MC.

−−→

MA

MA)

= 3 + 2(cos(−−→MB,−−→

MC) − cos(−−→MB,−−→

MA) − cos(−−→MC,−−→MA))

= 3 − 2(cos A + cos B + cos C) (Because M in small arc BC)

= 3 − 2R + r

R (Here we use the well-know equality: cos A + cos B + cos C =

R + r

R )

= R

2

− 2Rr

R2

= OI

2

R2 (Here we use the Euler’s formula)

From this we have

MO|

−−→

MB

MB +

−−→

MC

MC −

−−→

MA

MA| = OI (3) Thus from (1), (2), (3) we have the inequality MA + 2OI ≥ MB + MC ≥ MA − 2OI and equality when ABC is equaliteral triangle We completed the solution

Problem 2 Given are4ABC orthorcenter H, circumradius R, with any M on plane find minimum value of sum

MA3

+ MB3

+ MC3

−32R · MH2

Proof By AM-GM inequality we have

MA3

R2

+ MA2

3

R + R.MA ≥ 2MA2

⇒ MA

3

R ≥ 32MA2

− R

2

2 Similar we have

MB3

R ≥ 3

2MB

2

− R

2

2 ,

MC3

R ≥ 3

2MC

2

− R

2

2 Thus

MA3

+ MB3

+ MC3

2(MA

2

+ MB2

+ MC2

) −3

2R

2

(1) Called O is circumcenter of ABC

MA2

+ MB2

+ MC2

= (−−→MO +−→OA)2

+ (−−→MO +−−→OB)2

+ (−−→MO +−→OC)2

= 3MO2

+ 2−−→MO(−→OA +−−→OB +−→OC) + 3R2

= 3MO2

+ 2−−→

MO.−−→

OH + 3R2

= 3MO2

− (OM2

+ OH2

− MH2

) + 3R2

= 2MO2

− OH2

+ MH2

+ 3R2

≥ 3R2

− OH2

+ MH2

(2)

(Here we use familiar equal of vector −→OA +−−→OB +−→OC =−−→OH)

Form (1), (2) we have

MA3

+ MB+

MC3

R ≥ 32(3R2

− OH2

+ MH2

) − 32R2

⇒ MA3

+ MB3

+ MC3

− 32R.MH2

≥ 3R3

− 32R.OH2

= const Easily seen equal when M ≡ O

Thus we have MA3

+ MB3

+ MC3

− 32R.MH2

has minimum value iff M ≡ O

Problem 3 Given are 4ABC with sides a, b, c and 4A0B0C0 with sides a0, b0, c0 and area S0 With any M on plane prove that

a02

a MA +

b02

b MB +

c02

c MC ≥ 4S0

Proof We well know the inequality:

Given triangle ABC and ∀x, y, z > 0 then

(x + y + z)2

4 ≥ yz sin2

A + zx sin2

B + xy sin2

C

we can replace yz → x, zx → y, xy → z we will get the inequality:

1

4(

r yz

x +

r zx

y +

r xy

z )

2

≥ x sin2

A + y sin2

B + z sin2

C(∗) Now we let

x = MBMC bc.a02 = MBMC

bc.4R02sin2

A0

y = MCMA ca.b02 = MCMA

ca.4R02sin2

B0

z = MAMB ab.c02 = MAMB

ab.4R02sin2

C0

Thus we will have:

r yz

x =

v u u u u t

MCMA ca.b02 MAMB

ab.c02

x = MBMC bc.a02

= a

0

a.b0c0MA

similar we have

r zx

y =

b0

b.c0a0MB,r xy

z =

c0

c.a0b0MC and using inequality (∗) for triangle A0B0C0 and x, y, z as above we will get

1

4( X

a,b,c

a0

a.b0c0MA)2

a,b,c

MBMC bc.4R02sin2

A0 sin2

A0 = 1 4R02(X

a,b,c

MBMC

bc )

If we use well know inequality

MB.MC

MC.MA

MA.MB

ab ≥ 1 then we get the consequence inequality:

1

4( X

a,b,c

a0

a.b0c0MA)2

≥ 4R102

0

a.b0c0MA + b

0

b.c0a0MB + c

0

c.a0b0MC ≥ R10

⇔ a

02

a MA +

b02

b MB +

c02

c MC ≥ a

0b0c0

R0 = 4S0

Easily seen we have equal when 4A0B0C0 ∼ 4ABC and M ≡ H(Orthocenter of triangle ABC)

Remark This inequality have some nice applycations

• If 4A0B0C0 ≡ 4ABC we get the well know inequality aMA + bMB + cMC ≥ 4S

• If 4A0B0C0 ≡ 4JaJbJc with Ja, Jb, Jc are three excenter of triangle ABC with noitice a0 = 4R cosA

2, b

0 = 4R cosB

2, c

0 = 4R cosC

2 and S

0 = 2SR

r we will get the nice inequality cot A

2MA + cot

B

2MB + cot

C

2MC ≥ a + b + c

• If 4A0B0C0 ≡ 4BCA we will get the non symmetry inequality

b2

aMA +

c2

b MB +

a2

c MA ≥ 4S There are some nice other inequality is a consequence of this inequatlity

Problem 4 Let M be an arbitrary point inside equaliteral triangle ABC.Find min value of

1

MA+

1

MB +

1 MC Proof We can assume ABC is an equaliteral triangle with side 1,let barycentric coordinates of M

is (x, y, z), x + y + z = 1 because M inside triangle ⇒ x, y, z > 0 By distance formula we have

MA2

= y + z

x + y + za

2

− yz + zx + xy(x + y + z)2 a2

= by x + y + z = 1 and a = 1 ⇒ MA = py2+ yz + z2, similarly MB =√

z2 + zx + x2, MC =px2+ xy + y2 therefore we need find min value of

1

py2+ yz + z2 +√ 1

z2 + zx + x2 + 1

px2+ xy + y2

when x, y, z > 0, x + y + z = 1, we will solve it with Lagrange multipliers

WLOG 0 ≤ x ≤ y ≤ z < 1

Case 1: x = 0 we have to prove

f (y, z) = 1

py2+ yz + z2 + 1

y +

1

z ≥ 4 + 2

√ 3 3 indeed

f (y, z) ≥ f(√yz,√

yz) ≥ f(y + z

2 ,

y + z

2 ) = 4 +

2√ 3 3 Case 2: 0 < x ≤ y ≤ z < 1

F (x, y, z, λ) =X 1

px2+ xy + y2 + λ(x + y + z − 1)

∂x = 0 ,

∂F

∂y = 0 ,

∂F

∂z = 0 , then

−12

"

2x + y p(x2+ xy + y2)3 + 2x + z

p(z2+ zx + x2)3

# + λ = 0

−1 2

"

2y + x p(x2+ xy + y2)3 + 2y + z

p(y2+ yz + z2)3

# + λ = 0

−1 2

"

2z + x p(z2+ zx + x2)3 + 2z + y

p(y2+ yz + z2)3

# + λ = 0 Adding

λ = 1 2

"

x + y p(x2+ xy + y2)3 + y + z

p(y2+ yz + z2)3 + z + x

p(z2+ zx + x2)3

#

Inserting λ in first we get

p(x2+ xy + y2)3 = 1

p(y2+ yz + z2)3

Similarly

p(x2+ xy + y2)3 = 1

p(z2+ zx + x2)3

p(x2+ xy + y2)3 = 1

p(x2+ xy + z2)3

Hence

y2

(z2

+ zx + x2

)3

= z2

(x2

+ xy + y2

)3

We put y = ax, z = bx, where 1 ≤ a ≤ b

a2

(b2

+ b + 1)3

= b2

(a2

+ a + 1)3

we get a = b Hence y = z as necessary for critical points in the interior of the region 0 < x, y, z < 1

We have to prove

2

px2+ xy + y2 + 1

y√

3 ≥ 4 + 2

√ 3 3 where x + 2y = 1

p3y2

− 3y + 1 +

1

y√

3 ≥ 4 +2

√ 3 3 where

1

3 ≤ y ≤ 12

since x ≤ y ≤ z By differentiation it is easily checked that the absolute minimum of g(y) on  13,1

2



is 4 + 2

√

3

3 = g(1/2).

Thus 1

MA +

1

MB +

1

MC get min value ⇔ M(1

2,

1

2, 0)and others permutation ⇔ M concur three midpoint of three side

Problem 5 Suppose a, b, c are sidelengths of a triangle and ma, mb, mc are its medians Prove the inequality

ma

a2 + mb

b2 + mc

c2 ≥

√ 3(a2

+ b2

+ c2

) 2abc Proof This inequality equivalent

(mabc

a +

mbca

b +

mcab

c )

2

≥ 34(a2

+ b2

+ c2

)2

we have

(mabc

a +

mbca

b +

mcab

c )

2

≥ 3(X(mbca).(mbc cab)) = 3(Xa2

mbmc)

we will prove

3(Xa2

mbmc) ≥ 34(a2

+ b2

+ c2

)2

⇔ 4(Xa2

mbmc) ≥ (a2

+ b2

+ c2

)2

Indeed turn into triangle with three side ma, mb, mc we need prove:

X

4m2 a

3

4b

3

4c ≥ (m2

a+ m2

b + m2

c)2

⇔X(2(b2

+ c2

) − a2

)bc ≥ (a2

+ b2

+ c2

)2

by equivalent tranformation we have

⇔X12(a2

− (b − c)2

)(b − c)2

≥ 0 which is true because a > |b − c|, b > |c − a|, c > |a − b| with any triangle ABC

Problem 6 Let triangle ABC and X, Y, Z are arbitrary points on segment BC, CA, AB Prove that

1

SAY Z

SBZX

SCXY ≥ S 3

XY Z

Lemma 6.1 Let a, b, c > 0 be positive real numbers then

1 a(1 + b) +

1 b(1 + c) +

1 c(1 + a) ≥ 1 + abc3 Proof We have

(1 + abc)( 1

a(1 + b) +

1 b(1 + c) +

1 c(1 + a)) + 3 =

X 1 + a a(1 + b)+

Xb(c + 1)

1 + b ≥ √33

abc + 3

3

√ abc ≥ 6

So we are done.In fact the ineq could be better and stronger as

1 a(1 + b)+

1 b(1 + c)+

1 c(1 + a) ≥ √3 3

abc(1 +√3

abc)

Proof We let BX

BC = x,

CY

CA = y,

AZ

AB = z, SABC = S, 0 < x, y, z < 1 we will have

SAY Z

S = z(1 − y),SBZX

S = x(1 − z),SCXY

S = y(1 − x) Thus we have

SXY Z = S −SAY Z−SBZX−SCXY = S −S(z(1−y)+x(1−z)+y(1−x)) = S(xyz−(x−1)(y−1)(x−1)) Thus we need to prove

1

SAY Z

SBZX

SCXY ≥ 3

SXY Z

z(1 − y)+

S x(1 − z) +

S y(1 − x) ≥

3S xyz + (1 − x)(1 − y)(1 − z)

⇔ xy (1 − y)+

yz (1 − z) +

zx (1 − x) ≥

3

1 + (1 − x)(1 − y)(1 − z)

xyz Now let 1 − x

x = a > 0,

1 − y

y = b > 0,

1 − z

z = c > 0 we will get

1 (a + 1)b+

1 (b + 1)c +

1 (c + 1)a ≥ 3

1 + abc now replace a → b, b → c, c → a we will get our above lemma

Problem 7 Given two triangles ABC and A0B0C0 with ares S, S0 resp prove that

aa0+ bb0+ bb0 ≥ 4√3SS0

Proof

(Xaa0)2

≥ 3(Xbb0cc0) = 12SS0(X 1

sin A sin A0) = 24SS0X 1

cos(A − A0) − cos(A + A0) ≥

≥ 24SS0X 1

1 − cos(A + A0) = 12SS

0X 1 sin2 A + A0

2

≥ 48SS0 ⇒ aa0+ bb0+ cc0 ≥ 4√3SS0

In the last inequality we easily seen P A + A0

2 = π thus they are three angle of a triangle therefore apply the inequality P 1

sin2

A ≥ 4 for them,we are done