mr 2 2015 solutions

Junior problems

J331. Determine all positive integers n such that

n!

(1 +

1

2+ · · ·+ 1

n+

1

n!

)is divisible by n.

Proposed by Alessandro Ventullo, Milan, Italy

Solution by Henry Ricardo, New York Math CircleWe get divisibility by n for n = 1 and for n a prime.

Clearly the given product is divisible by n when n = 1. For n ≥ 2 we have

n!

(1 +

1

2+ · · ·+ 1

n+

1

n!

)= n!

(1 +

1

2+ · · ·+ 1

n− 1

)+ (n− 1)! + 1

≡ (n− 1)! + 1 (mod n) ≡ 0 (mod n)⇐⇒ n is prime

by Wilson’s theorem.

Also solved by Daniel Lasaosa, Pamplona, Spain; Brian Bradie, Christopher Newport University, NewportNews, VA, USA; Sardor Bozorboyev, S.H.Sirojjidinov lyceum, Tashkent, Uzbekistan; Haimoshri Das, SouthPoint High School, India; Adnan Ali, A.E.C.S-4, Mumbai, India; Alberto Espuny Díaz, CFIS, UniversitatPolitècnica de Catalunya, Barcelona, Spain; Arkady Alt, San Jose, CA, USA; Corneliu Mănescu-Avram,Transportation High School, Ploieşti, Romania; Byeong Yeon Ryu, The Hotchkiss School, Lakeville, CT,USA; David E. Manes, Oneonta, NY, USA; Problem Solving Group, Department of Financial and Manage-ment Engineering, University of the Aegean, Greece; G. C. Greubel, Newport News, VA, USA; James Pierog,SMIC School, Shanghai, China; Robert Bosch, Archimedean Academy, FL, USA; Albert Stadler, Herrliberg,Switzerland; Suhas Vadan Gondi, KV-IISc High School, Bengaluru, India; Neculai Stanciu, George EmilPalade School, Buzău, Romania and Titu Zvonaru, Comăneşti, Romania; Francesco Bonesi, University ofEast Anglia, UK; Polyahedra, Polk State College, FL, USA; Kwonil Kobe Ko, Cushing Academy, MA, USA;George Gavrilopoulos, Nea Makri High School, Athens, Greece; AN-anduud Problem Solving Group, Ulaan-baatar, Mongolia; Dhruv Nevatia, Ramanujan Academy, Nashik, Maharashtra, India; Utamuratov Odilbek,Academic Lyceum Nr.2, Uzbekistan; Michael Tang, Edina High School, MN, USA; Mahmoud Ezzaki andChakib Belgani; Louis Cahyadi, Cirebon, Indonesia; Daniel Jhiseung Hahn, Phillips Exeter Academy, Exe-ter, NH, USA; Théo Lenoir, Institut Saint-Lô, Agneaux, France; Haroun Meghaichi, University of Scienceand Technology Houari Boumediene, Algiers, Algeria.

Mathematical Reflections 2 (2015) 1

J332. Let n ≥ 3 and 0 = a0 < a1 < · · · < an+1 such that a1a2 + a2a3 + · · ·+ an−1an = anan+1. Prove that

1

a32 − a02+

1

a42 − a12+ · · ·+ 1

an+12 − an−22

≥ 1

an−12.

Proposed by Titu Andreescu, University of Texas at Dallas

Solution by Brian Bradie, Christopher Newport University, Newport News, VA, USAThe expression on the left-hand side can be rewritten as

a21a22

a21a22a

23 − a20a21a22

+a22a

23

a22a23a

24 − a21a22a23

+ · · ·+a2n−1a

2n

a2n−1a2na

2n+1 − a2n−2a2n−1a2n

.

Applying the Cauchy-Schwartz inequality then yields

a21a22

a21a22a

23 − a20a21a22

+a22a

23

a22a23a

24 − a21a22a23

+ · · ·+a2n−1a

2n

a2n−1a2na

2n+1 − a2n−2a2n−1a2n

≥ (a1a2 + a2a3 + · · ·+ an−1an)2

a21a22a

23 − a20a21a22 + a22a

23a

24 − a21a22a23 + · · ·+ a2n−1a

2na

2n+1 − a2n−2a2n−1a2n

=a2na

2n+1

a2n−1a2na

2n+1 − a20a21a22

=1

a2n−1.

Also solved by Daniel Lasaosa, Pamplona, Spain; Polyahedra, Polk State College, FL, USA; KwonilKobe Ko, Cushing Academy, MA, USA; George Gavrilopoulos, Nea Makri High School, Athens, Greece; AN-anduud Problem Solving Group, Ulaanbaatar, Mongolia; Daniel Jhiseung Hahn, Phillips Exeter Academy,Exeter, NH, USA; Sardor Bozorboyev, S.H.Sirojjidinov lyceum, Tashkent, Uzbekistan; Adnan Ali, A.E.C.S-4,Mumbai, India; Byeong Yeon Ryu, The Hotchkiss School, Lakeville, CT, USA; Nicusor Zlota, Traian VuiaTechnical College, Focsani, Romania; Paolo Perfetti, Università degli studi di Tor Vergata Roma, Roma,Italy.


J333. Consider an equiangular hexagon ABCDEF . Prove that

AC2 + CE2 + EA2 = BD2 +DF 2 + FB2.

Proposed by Nairi Sedrakyan, Armenia

Solution by Robert Bosch, Archimedean Academy, Florida, USADenote the sides of equiangular hexagon by a, b, c, d, e, f in clockwise order. More precisely,

AB = a, BC = b, CD = c, DE = d, EF = e, FA = f.

The following lemma is true.a− d = e− b = c− f.

The proof may be found in the paper “Equiangular polygons. An algebraic approach. Titu Andreescu, Bog-dan Enescu. Mathematical Reflections.

Now, by Cosines’s law, it follows

AC2 = a2 + b2 − 2ab cosα,

CE2 = c2 + d2 − 2cd cosα,

EA2 = e2 + f2 − 2ef cosα,

BD2 = b2 + c2 − 2bc cosα,

DF 2 = d2 + e2 − 2de cosα,

FB2 = f2 + a2 − 2fa cosα.

Where, α is the common inner angle. By above relations,

AC2 + CE2 + EA2 = BD2 +DF 2 + FB2 ⇔ ab+ cd+ ef = bc+ de+ fa.

By lemma:

2(ab− de) = e2 + d2 − a2 − b2,2(cd− fa) = a2 + f2 − c2 − d2,2(ef − bc) = b2 + c2 − e2 − f2.

Adding, we get the conclusion.

Also solved by Daniel Lasaosa, Pamplona, Spain; Polyahedra, Polk State College, FL, USA; Kwonil KobeKo, Cushing Academy, MA, USA; Dhruv Nevatia, Ramanujan Academy, Nashik, Maharashtra, India; Uta-muratov Odilbek, Academic Lyceum Nr.2, Uzbekistan; Michael Tang, Edina High School, MN, USA; DanielJhiseung Hahn, Phillips Exeter Academy, Exeter, NH, USA; Sardor Bozorboyev, S.H.Sirojjidinov lyceum,Tashkent, Uzbekistan; Neculai Stanciu, George Emil Palade School, Buzău, Romania and Titu Zvonaru, Co-măneşti, Romania; Byeong Yeon Ryu, The Hotchkiss School, Lakeville, CT, USA; Corneliu Mănescu-Avram,Transportation High School, Ploieşti, Romania; David E. Manes, Oneonta, NY, USA; Nicolas Huynh, LycéeRichelieu, Rueil-Malmaison, France; Nicusor Zlota, Traian Vuia Technical College, Focsani, Romania.


J334. Let ABC be a triangle with ∠A ≥ 60◦ and let D and E be points on the sides AB and AC, respectively.Prove that

BC

min (BD,DE,EC)≥√

5− 4 cosA.

Proposed by Ivan Borsenco, Massachusetts Institute of Technology, USA

Solution by the author Suppose BD = min (BD,DE,EC). We can start moving segment BD closerto A keeping the distance BD fixed. Say this achieved for B′D′ = BD, where B′ and D′ are new points.We draw a line through D′ parallel to DE and a line through B′ parallel to BC. Because AB′ < AB, wehave D′E′ < DE and B′C ′ < BC. Using this optimizing procedure can reach a configuration with eitherBD = DE = min(BD,DE,EC) or BD = EC = min(BD,DE,EC), because min (BD,DE,EC) staysfixed, while BC is decreasing.

Suppose DE = min (BD,DE,EC). Then we can can draw a line B′C ′ closer to A and parallel to BCsuch that B′D < BD and C ′E < CE. Using this optmizing procedure, without loss of generality, we reacha configuration with BD = DE = min(BD,DE,EC).

Consider the configuration with BD = EC = min(BD,DE,EC). Denote by u = AD, v = AE,x = BD = CE and y = DE, where y ≥ x. Then

a2 = (u+ x)2 + (v + x)2 − 2(u+ x)(v + x) cosA, y2 = u2 + v2 − 2uv cosA.

It follows thata2 = y2 + 2x2(1− cosA) + 2(u+ v)x(1− cosA) ≥ x2(5− 4 cosA),

because y ≥ x and (u+ v) ≥ x.Consider the configuration with BD = DE = min(BD,DE,EC). We can attempt to move point C,

while minimize EC and BC. This is possible if ∠C < 90◦. However, if ∠C ≥ 90◦ it is not, because BC isnot going to lessen. From here, we can reach two possible configurations:• BD = DE = EC, which follows from the previously analyzed case.• BD = DE < CE and ∠C ≥ 90◦.The distance from D to BC is x sinB and the distance from E to BC is y sin(A+B). Because y sin(A+

B) ≥ x sinB, we get ∠DEA is obtuse. Hence DC ≤ DA and x ≤ AB2 and also x ≤ y ≤ AC. Therefore

min(BD,DE,EC) ≤ min(c2 , b).

a2 = b2 + c2 − 2bc cosA = 4( c

2

)2+ b2 − 4b

( c2

)cosA ≥

≥ 3( c

2

)2+ 2b

( c2

)(1− 2 cosA) ≥ (5− 4 cosA)(min(BD,DE,EC))2,

as desired.


J335. Prove that for any a, b > −1,

max{

(a+ 3)(b2 + 3

), (b+ 3)

(a2 + 3

)}≥ 2(a+ b+ 2)

32 .


Solution by Michael Tang, Edina High School, MN, USASince a+ 1, b+ 1 > 0, by Hölder’s Inequality,

(1 + 1)((a+ 1)3 + 23)(23 + (b+ 1)3) ≥ (2(a+ 1) + 2(b+ 1))3

2((a+ 1)3 + 8)((b+ 1)3 + 8) ≥ 8(a+ b+ 2)3

((a+ 1)3 + 8)((b+ 1)3 + 8) ≥ 4(a+ b+ 2)3.

But note that (a+ 1)3 + 8 = a3 + 3a2 + 3a+ 9 = (a+ 3)(a2 + 3), so we get

(a+ 3)(a2 + 3)(b+ 3)(b2 + 3) ≥ 4(a+ b+ 2)3.

Now letX = max{(a+ 3)(b2 + 3), (a2 + 3)(b+ 3)}.

Then X ≥ (a+ 3)(b2 + 3) and X ≥ (a2 + 3)(b+ 3), so

X2 ≥ (a+ 3)(a2 + 3)(b+ 3)(b2 + 3) ≥ 4(a+ b+ 2)3.

Thus X ≥ 2(a+ b+ 2)3/2.

Also solved by Arkady Alt, San Jose, CA, USA; Polyahedra, Polk State College, USA; Kwonil Kobe Ko,Cushing Academy, MA, USA; Mahmoud Ezzaki and Chakib Belgani; Daniel Jhiseung Hahn, Phillips ExeterAcademy, Exeter, NH, USA; Sardor Bozorboyev, S.H.Sirojjidinov lyceum, Tashkent, Uzbekistan; ByeongYeon Ryu, The Hotchkiss School, Lakeville, CT, USA; Haimoshri Das, South Point High School, India;Nicusor Zlota, Traian Vuia Technical College, Focsani, Romania; Paolo Perfetti, Università degli studi diTor Vergata Roma, Roma, Italy; Inequaliter and Equaliter; Neculai Stanciu, George Emil Palade School,Buzău, Romania and Titu Zvonaru, Comăneşti, Romania.


J336. Let ABC be a right triangle with altitude AD, and let T be an arbitrary point on segment BD. Linesthrough T tangent to the circumcircle of triangle ADC intersect line AB at X,Y , respectively. Provethat AX = BY .

Proposed by Josef Tkadlec, Charles University, Czech Republic

Solution by Polyahedra, Polk State College, USA

Z

B

M

L

I

X

Y

NC

O

A

D

T

In the figure, CZ ‖ AB. Let k = AX/CZ and let s be the semiperimeter of 4TXY . Then LT =s − XY = AY − XY = AX = kNZ, and Y L = XM tanα/ tanβ = (s − Y T )k = k(Y N − Y T ) = kTN .Hence,

BY

CZ=Y T

TZ=

Y L+ LT

TN +NZ= k,

thus BY = AX.

Also solved by George Gavrilopoulos, Nea Makri High School, Athens, Greece; Michael Tang, Edina HighSchool, MN, USA; Arkady Alt, San Jose, CA, USA; Robert Bosch, Archimedean Academy, FL, USA; NeculaiStanciu, George Emil Palade School, Buzău, Romania and Titu Zvonaru, Comăneşti, Romania.


Senior problems

S331. Find the minimum value of

E (x1, . . . , xn) = x12 + · · ·+ xn

2 + x1x2 + · · ·+ x1xn + · · ·+ xn−1xn + x1 + · · ·+ xn,

when x1, . . . , xn ∈ R.

Proposed by Dorin Andrica, Babeş-Bolyai University, Romania

Solution by Daniel Lasaosa, Pamplona, SpainDenote s = x1 + x2 + · · ·+ xn. We can then rewrite

E (x1, . . . , xn) =s2

2+

(x1 + 1)2 + · · ·+ (xn + 1)2

2− n

2≥

≥ s2

2+n

2

((x1 + 1) + · · ·+ (xn + 1)

n

)2

− n

2=

(n+ 1)s2 + 2ns

2n=

=((n+ 1)s+ n)2

2n(n+ 1)− n

2(n+ 1)≥ − n

2(n+ 1).

The first inequality is a consequence of the AM-QM inequality, or equality holds iff x1 = · · · = xn, whereasequality holds in the second equality iff s = − n

n+1 . It follows that the minimum value of E (x1, . . . , xn) is− n

2(n+1) , obtained iff x1 = · · · = xn = − 1n+1 .

An alternative (less Olympic) method arises from considering that the first and second derivatives ofE (x1, . . . , xn) with respect to xi for i = 1, . . . , n are respectively s+xi + 1 and 2, whereas any mixed secondderivative is 1 (resulting in a positive-definite Hessian matrix) or a local minimum is reached wheneversn = x1 = · · · = xn = −s− 1, resulting again in the minimum of E (x1, . . . , xn) occurring at x1 = · · · = xn =− 1

n+1 , and reaching the same value − n2(n+1) .

Also solved by Brian Bradie, Christopher Newport University, Newport News, VA, USA; Kwonil Ko-be Ko, Cushing Academy, MA, USA; AN-anduud Problem Solving Group, Ulaanbaatar, Mongolia; MichaelTang, Edina High School, MN, USA; Mahmoud Ezzaki and Chakib Belgani; Théo Lenoir, Institut Saint-Lô,Agneaux, France; Sardor Bozorboyev, S.H.Sirojjidinov lyceum, Tashkent, Uzbekistan; Adnan Ali, A.E.C.S-4,Mumbai, India; Byeong Yeon Ryu, The Hotchkiss School, Lakeville, CT, USA; Moubinool Omarjee LycéeHenri IV, Paris France; Paolo Perfetti, Università degli studi di Tor Vergata Roma, Roma, Italy; Albert Stad-ler, Herrliberg, Switzerland; Francesco Bonesi, University of East Anglia, UK; Robert Bosch, ArchimedeanAcademy, FL, USA.


S332. Prove that in any triangle with side lengths a, b, c and median lengths ma,mb,mc,

4 (ma +mb +mc) ≤√

8a2 + (b+ c)2 +√

8b2 + (c+ a)2 +√

8c2 + (a+ b)2.

Proposed by Titu Andreescu, University of Texas at Dallas, USA

Solution by AN-anduud Problem Solving Group, Ulaanbaatar, MongoliaWe have

(a− b)2(a+ b+ c)(a+ b− c) ≥ 0

⇔ (a− b)2((a+ b)2 − c2) ≥ 0

⇔ (a2 − b2)2 ≥ (ac− bc)2

⇔ 4c2ab ≥ 2(ac)2 + 2(bc)2 − 2(a2 − b2)2

⇔ 4c4 + 4c2ab+ (ab)2 ≥ 2(ac)2 + 2(bc)2 + 4c4 + (ab)2 − 2(a2 − b2)2

⇔ (2c2 + ab)2 ≥ (2b2 + 2c2 − a2)(2c2 + 2a2 − b2)

⇔ 2(2c2 + ab) ≥ 2√

(2b2 + 2c2 − a2)(2c2 + 2a2 − b2)

⇔ 8c2 + a2 + b2 + 2ab ≥ 2√

(2b2 + 2c2 − a2)(2c2 + 2a2 − b2) + 4c2 + a2 + b2

⇔√

8c2 + (a+ b)2 ≥√

2c2 + 2b2 − a2 +√

2c2 + 2a2 − b2.

Similarly we obtain √8b2 + (c+ a)2 ≥

√2c2 + 2b2 − a2 +

√2a2 + 2b2 − c2,√

8a2 + (c+ b)2 ≥√

2a2 + 2b2 − c2 +√

2c2 + 2a2 − b2.

Adding these inequalities and using√2c2 + 2b2 − a2 = 2ma,

√2c2 + 2a2 − b2 = 2mb,

√2a2 + 2b2 − c2 = 2mc

we get the required inequality. Equality holds if and only if the triangle is equilateral.

Also solved by Daniel Lasaosa, Pamplona, Spain; José Hernández Santiago, México; Kwonil Kobe Ko,Cushing Academy, MA, USA; Louis Cahyadi, Cirebon, Indonesia; Inequaliter and Equaliter; Sardor Bozor-boyev, S.H.Sirojjidinov lyceum, Tashkent, Uzbekistan; Arkady Alt, San Jose, CA, USA; Byeong Yeon Ryu,The Hotchkiss School, Lakeville, CT, USA; Nicusor Zlota, Traian Vuia Technical College, Focsani, Romania;Rade Krenkov, SOU Goce Delcev, Valandovo, Macedonia; Robert Bosch, Archimedean Academy, FL, USA;Neculai Stanciu, George Emil Palade School, Buzău, Romania and Titu Zvonaru, Comăneşti, Romania.


S333. Let x > 1 and let (an)n≥1 be the sequence defined by an = [xn] for every positive integer n. Prove thatif (an)n≥1 is a geometric progression, then x is an integer.

Proposed by Marius Cavachi, Constanta, Romania

Solution by Robert Bosch, Archimedean Academy, Florida, USADenote by k the ratio of the geometric progression. We have bxnc = kn−1bxc, now we fix x and dividing by

xn both sides we getbxncxn

=

(k

x

)n−1 bxcx

. Now we have three different cases, say 0 <k

x< 1,

k

x= 1 or

k

x> 1. Note that

xn − 1

xn<bxncxn

< 1, thus limn→∞

bxncxn

= 1. So, taking the limit when n tends to infinity we

reach to a contradiction unlessk

x= 1, thus getting bxc = x, which means x must be an integer.

Also solved by Daniel Lasaosa, Pamplona, Spain; Byeong Yeon Ryu, The Hotchkiss School, Lakeville,CT, USA; Alberto Espuny Díaz, CFIS, Universitat Politècnica de Catalunya, Barcelona, Spain; Sardor Bo-zorboyev, S.H.Sirojjidinov lyceum, Tashkent, Uzbekistan; Théo Lenoir, Institut Saint-Lô, Agneaux, France;Mahmoud Ezzaki and Chakib Belgani; Kwonil Kobe Ko, Cushing Academy, MA, USA.


S334. Let a0 ≥ 0 and an+1 = a0 · . . . · an + 4 for n ≥ 0. Prove that

an − 4√

(an+1 + 1) (an2 + 1)− 4 = 1

for all n ≥ 1.


Solution by AN-anduud Problem Solving Group, Ulaanbaatar, MongoliaWe have

an+1 = a0 · . . . · an−1an + 4

⇒ an+1 = (an − 4)an + 4

⇒ an+1 + 1 = a2n − 4an + 5

⇒ (an+1 + 1)(a2n + 1)− 4 = (a2n − 4an + 5)(a2n + 1)− 4

⇒ (an+1 + 1)(a2n + 1)− 4 = a4n − 4a3n + 6a2n − 4an + 1

⇒ (an+1 + 1)(a2n + 1)− 4 = (an − 1)4

⇒√

(an+1 + 1)(a2n + 1)− 4 = an − 1

⇒ an − 4√

(an+1 + 1)(a2n + 1)− 4 = 4.

Also solved by Daniel Lasaosa, Pamplona, Spain; Brian Bradie, Christopher Newport University, New-port News, VA, USA; George Gavrilopoulos, Nea Makri High School, Athens, Greece; Mahmoud Ezzaki andChakib Belgani; Théo Lenoir, Institut Saint-Lô, Agneaux, France; Sardor Bozorboyev, S.H.Sirojjidinov ly-ceum, Tashkent, Uzbekistan; Adnan Ali, A.E.C.S-4, Mumbai, India; Arkady Alt, San Jose, CA, USA; ByeongYeon Ryu, The Hotchkiss School, Lakeville, CT, USA; David E. Manes, Oneonta, NY, USA; G. C. Greubel,Newport News, VA, USA; Haimoshri Das, South Point High School, India; Haroun Meghaichi, Universityof Science and Technology Houari Boumediene, Algiers, Algeria; Henry Ricardo, New York Math Circle;Moubinool Omarjee Lycée Henri IV, Paris France; Paolo Perfetti, Università degli studi di Tor VergataRoma, Roma, Italy; Rade Krenkov, SOU Goce Delcev, Valandovo, Macedonia; Robert Bosch, ArchimedeanAcademy, FL, USA; Russelle Guadalupe, University of the Philippines - Diliman, Quezon City, Philippines;Albert Stadler, Herrliberg, Switzerland; Neculai Stanciu, George Emil Palade School, Buzău, Romania andTitu Zvonaru, Comăneşti, Romania.


S335. Let ABC be a triangle. Let D be a point on the ray BA, which is not on the side AB, and let E bea point on the side AC, which is different from A And C. Let X be the reflection of D with respectto B and let Y be the reflection of E with respect to C. Suppose 4BC2 + DE2 = XY 2. Prove thatBE ⊥ CD if and only if ∠BAC = 90◦.

Proposed by İlker Can Çiçek, Instanbul, Turkey

Solution by Andrea Fanchini, Cantú, ItalyWe set AD = d and AE = e, then points D and E have barycentric coordinates D(c + d : −d : 0),

E(b− e : 0 : e).So, points X and Y will have coordinates

X(−c− d : 2c+ d : 0), Y (e− b : 0 : 2b− e)

with the usual distance formula and the Conway’s notation we have

DE2 =SA(ce+ bd)2 + SBb

2d2 + SCc2e2

b2c2=bcd2 + bce2 + 2edSA

bc

XY 2 =SA((2b− e)(−c− d) + (b− e)(2c+ d))2 + SB(b(2c+ d))2 + SC(c(e− 2b))2

b2c2=

=bcd2 + bce2 + 4a2bc+ 2edSA + 4bdSB − 4ceSC

bc

with the statement 4BC2 +DE2 = XY 2, it follows that

4a2 +bcd2 + bce2 + 2edSA

bc=bcd2 + bce2 + 4a2bc+ 2edSA + 4bdSB − 4ceSC

bc

developing and simplifying, we obtain that

bdSB = ceSC (∗)

I) ∠BAC = 90◦ ⇒ BE ⊥ CD

If ∠BAC = 90◦ then SA = 0 and SB = c2, SC = b2. Then it follows that the (∗) relation it becomes

bdSB = ceSC ⇒ cd = be

the line BE have equation ex+ (e− b)z = 0 and the infinite point is BE∞(b− e : −b : e),

the line CD have equation dx+ (c+ d)y = 0 and the infinite perpendicular point is

CD∞⊥(da2 − (c+ d)SC : b2(c+ d)− dSC : −dSB − (c+ d)SA)

but SA = 0, SB = c2, SC = b2 and cd = be so developing and simplifying

CD∞⊥ ≡ (b− e : −b : e) = BE∞ ⇒ BE ⊥ CD, q.e.d.

II) BE ⊥ CD ⇒ ∠BAC = 90◦


now we know that two lines pix+ qiy + riz = 0(i = 1, 2) are perpendiculars if and only if

SA(q1 − r1)(q2 − r2) + SB(r1 − p1)(r2 − p2) + SC(p1 − q1)(p2 − q2) = 0

so for the lines BE and CD we have

SA(b− e)(c+ d) + bdSB − ceSC = 0

but substituing the (∗) relation it follows that SA = 0 and this means that ∠BAC = 90◦, q.e.d.

Also solved by Daniel Lasaosa, Pamplona, Spain; Sardor Bozorboyev, S.H.Sirojjidinov lyceum, Tashkent,Uzbekistan; Byeong Yeon Ryu, The Hotchkiss School, Lakeville, CT, USA; Neculai Stanciu, George EmilPalade School, Buzău, Romania and Titu Zvonaru, Comăneşti, Romania.


S336. Let M be a point inside the triangle ABC and let K be a point symmetric to point M with respectto AC. Line BM intersects AC in point N and line BK intersects AC in point P . Prove that if∠AMP = ∠CMN , then ∠ABP = ∠CBN .


Solution by Daniel Lasaosa, Pamplona, SpainConsider the circumcircle Γ of BMK, which is clearly symmetric with respect to AC. By construction, MPis the symmetric of BK, and NK is the symmetric of BM , or they meet again at a point B′ on Γ whichis the symmetric of B with respect to AC. Let Q be the point where segment AC intersects Γ, or clearlyBQ = BQ′, and

∠PMQ = ∠QMB′ = ∠BKQ = 180◦ − ∠BMQ = ∠QMN,

or ∠AMQ = ∠CMQ. Now, applying the Sine Law to triangles AMQ and CMQ, and noting thatsin∠AQM = sin∠CQM because ∠AQM + ∠CQM = 180◦, we conclude that AM

AQ = CMCQ , or M is on

the Apollonius’ circle defined by AXCX = AB

CB since the circle through every pointM and Q passes also throughB. We conclude that if ∠AMP = ∠CMN , then M is necessarily on the circle symmetric with respect toAC, through B, and through the intersection of AC and the internal bisector of angle B.

Since BQ is then the internal bisector of angle B, it follows that ∠CBQ = ∠ABQ, or it suffices to provethat ∠MQB = ∠QBK, clearly true since MQ = QK by symmetry with respect to AC, and BMQK iscyclic. The conclusion follows.

Also solved by Arkady Alt, San Jose, CA, USA; Sardor Bozorboyev, S.H.Sirojjidinov lyceum, Tashkent,Uzbekistan; Haimoshri Das, South Point High School, India.


Undergraduate problems

U331. Find all positive integers a > b ≥ 2 such that

ab − a = ba − b.

Proposed by Mircea Becheanu, Bucharest, Romania

Solution by Robert Bosch, Archimedean Academy, FL, USATo start, consider b = 2; we have to solve in positive integers the equation a2 − a + 2 = 2a with a ≥ 3.The inequality 2a > a2 for a ≥ 5 is easily provable by induction. Hence by the above equation we obtaina2 − a + 2 > a2, which it holds only when a < 2. Then, we must consider the values a = 3 and a = 4,

which leads to the solution (a, b) = (3, 2). When a > b > 2, we consider the function f(x) =lnx

x, which is

decreasing when x > e because the derivative f ′(x) =1− lnx

x2is clearly negative. Now we have the following

chain of equivalences

a > b↔ ln a

a<

ln b

b↔ ab < ba ↔ ab − ba < 0↔ a− b < 0↔ a < b.

Contradiction. Hence the only solution is (a, b) = (3, 2).

Also solved by Daniel Lasaosa, Pamplona, Spain; Francesco Bonesi, University of East Anglia, UK;David E. Manes, Oneonta, NY, USA; Byeong Yeon Ryu, The Hotchkiss School, Lakeville, CT, USA; SardorBozorboyev, S.H.Sirojjidinov lyceum, Tashkent, Uzbekistan; Théo Lenoir, Institut Saint-Lô, Agneaux, France;Mahmoud Ezzaki and Chakib Belgani.


U332. Find inf(x,y)∈D(x+ 1)(y + 1), where D = {(x, y)|x, y ∈ R+, x 6= y, xy = yx}.

Proposed by Arkady Alt, San Jose, USA

Solution by Daniel Lasaosa, Pamplona, SpainNote that ey lnx = xy = yx = ex ln y iff lnx

x = ln yy , since ex is a strictly increasing function for all real x. Note

also thatd

dx

(ln(x)

x

)=

1− ln(x)

x2

is negative iff x > e, positive iff x < e, and zero iff x = e, ie for any x 6= y such that xy = yx, we must haveeither x > e > y or x < e < y. We may therefore define D∗ = D ∪ {(x, y) = (e, e)}, and the problem isequivalent to finding, for (x+ 1)(y + 1), either its minimum in D∗ if it exists at (x, y) = (e, e) (since it willcoincide by continuity of functions (x+ 1)(y+ 1), xy, yx with the infimum in D), or otherwise its infimum inD.

Define f(x, y) = xy, and g(x, y) = x ln(y)−y ln(x). Note that the extrema of f(x, y) subject to conditiong(x, y) = 0 may be found by Lagrange’s multiplier method, or real constant λ exists such that

λy = ln(y)− y

x, λxy = x ln(y)− y = y ln(x)− x,

and since x ln(y) = y ln(x), we find that a local extremum occurs iff x = y, ie the only local extremum of xyin D∗ occurs when x = y = e, with a value e2. Moreover, the borders of D∗ occur when x→ 1 and y →∞or vice versa, for xy → ∞, or the extremum of xy at x = y = e is a minimum with value e2. We concludethat, for (x, y) ∈ D∗, we have

(x+ 1)(y + 1) = xy + x+ y + 1 ≥ xy + 2√xy + 1 ≥ e2 + 2e+ 1 = (e+ 1)2,

with equality iff x = y = e. By continuity of condition xy = yx and of function (x+ 1)(y + 1), we concludethat

inf(x,y)∈D

(x+ 1)(y + 1) = (e+ 1)2,

where (x + 1)(y + 1) can get arbitrarily close to (e + 1)2 as x > e gets arbitrarily close to e, with thecorresponding value of y < e getting arbitrarily close to e too, or vice versa.

Also solved by Francesco Bonesi, University of East Anglia, UK; Robert Bosch, Archimedean Academy,FL, USA; Paolo Perfetti, Università degli studi di Tor Vergata Roma, Roma, Italy; AN-anduud ProblemSolving Group, Ulaanbaatar, Mongolia.


U333. Evaluate ∏n≥0

(1− 22

n

22n+1 + 1

).

Proposed by Titu Andreescu, University of Texas at Dallas, USA

Solution by G.R.A.20 Problem Solving Group, Roma, ItalyLet xn = 22

n , then xn+1 = x2n, and∏n≥0

(1− 22

n

22n+1 + 1

)=∏n≥0

(1− 1/xn

1/x2n + 1

)

=∏n≥0

x2n − xn + 1

1 + x2n· 1 + xn

1 + xn· 1− xn

1− xn· 1− x3n

1− x3n

=∏n≥0

(1− x6n)(1− xn)

(1− x4n)(1− x3n)

=∏n≥0

1− x3n+1

1− x3n·∏n≥0

1− xn1− xn+2

=(1− x0)(1− x1)

1− x30=

(1− 1/2)(1− 1/4)

1− 1/8=

3

7.

Also solved by Albert Stadler, Herrliberg, Switzerland; Daniel Lasaosa, Pamplona, Spain; Robert Bosch,Archimedean Academy, FL, USA; Paolo Perfetti, Università degli studi di Tor Vergata Roma, Roma, Italy;Moubinool Omarjee Lycée Henri IV, Paris France; Byeong Yeon Ryu, The Hotchkiss School, Lakeville,CT, USA; Arkady Alt, San Jose, CA, USA; Ángel Plaza, University of Las Palmas de Gran Canaria,Spain; Adnan Ali, A.E.C.S-4, Mumbai, India; Michael Tang, Edina High School, MN, USA; AN-anduudProblem Solving Group, Ulaanbaatar, Mongolia; Kwonil Kobe Ko, Cushing Academy, MA, USA; BrianBradie, Christopher Newport University, Newport News, VA, USA.


U334. Prove that if x ∈ R with |x| ≥ e, then e|x| ≥(e2+x2

2e

)e, while the inequality is reversed if |x| ≤ e.

Proposed by Angel Plaza, Universidad de Las Palmas de Gran Canaria, Spain

Solution by Robert Bosch, Archimedean Academy, FL, USATaking natural logarithms on both sides and dividing by e, the inequality becomes∣∣∣x

e

∣∣∣ > ln

((xe

)2+ 1

)+ 1− ln 2,

so, after the substitution y =x

ewe must show that

|y| ≥ 1⇒ |y| > ln(y2 + 1) + 1− ln 2.

Now we consider two cases: when y ≥ 1 and when y ≤ −1. In the first one we consider the function

f(y) = ln(y2 + 1) − y + 1 − ln 2 with derivative f ′(y) =2y

y2 + 1− 1 < 0, hence f(y) is decreasing, it

follows that f(y) ≤ f(1) = 0. Now, in the second one g(y) = ln(y2 + 1) + y + 1 − ln 2 with first derivative

g′(y) =2y

y2 + 1+ 1 > 0, thus is an increasing function, and then g(y) ≤ g(−1) = 0. In a similar way the

reversed inequality may be proved.

Also solved by Daniel Lasaosa, Pamplona, Spain; Francesco Bonesi, University of East Anglia, UK; AlbertStadler, Herrliberg, Switzerland; Paolo Perfetti, Università degli studi di Tor Vergata Roma, Roma, Italy;Haroun Meghaichi, University of Science and Technology Houari Boumediene, Algiers, Algeria; Byeong YeonRyu, The Hotchkiss School, Lakeville, CT, USA; Corneliu Mănescu-Avram, Transportation High School, Plo-ieşti, Romania; Arkady Alt, San Jose, CA, USA; Théo Lenoir, Institut Saint-Lô, Agneaux, France; MahmoudEzzaki and Chakib Belgani; AN-anduud Problem Solving Group, Ulaanbaatar, Mongolia; Kwonil Kobe Ko,Cushing Academy, MA, USA; Brian Bradie, Christopher Newport University, Newport News, VA, USA.


U335. Let p, a1, ..., an, b1, ..., bn be positive real numbers. Prover that

a1

(a1b1

)p

+ ...+ an

(a1 + ...+ anb1 + ...+ bn

)p

<

(p+ 1

p

)p(a1p+1

b1p + ...+

anp+1

bnp

).


Solution by AN-anduud Problem Solving Group, Ulaanbaatar, MongoliaWe need the weighted Hardy inequality, which is stated as: Let a1, . . . , an and ω1, . . . , ωn be positive realnumbers. If p > 1 then

n∑k=1

ωk

(ω1a1 + · · ·+ ωkakω1 + · · ·+ ωk

)p

<

(p

p− 1

)p n∑k=1

ωkapk.

Replacing p, ak, ωk by p+ 1, akbk , bk for k = 1, . . . , n, the inequality reduces to

n∑k=1

bk

(a1 + · · ·+ akb1 + · · ·+ bk

)p+1

<

(p+ 1

p

)p+1 n∑k=1

ap+1k

bpk.

By Hölder’s inequality and the above inequality we have

n∑k=1

ak

(a1 + · · ·+ akb1 + · · ·+ bk

)p

=

n∑k=1

ak

bp

p+1

k

(b

pp+1

k

(a1 + · · ·+ akb1 + · · ·+ bk

)p)

≤

(n∑

k=1

ap+1k

bpk

) 1p+1(

n∑k=1

bk

(a1 + · · ·+ akb1 + · · ·+ bk

)p+1) p

p+1

<

(n∑

k=1

ap+1k

bpk

) 1p+1((

p+ 1

p

)p+1 n∑k=1

ap+1k

bpk

) pp+1

=

(p

p+ 1

)p n∑k=1

ap+1k

bpk,

and the inequality is proved.

Also solved by Robert Bosch, Archimedean Academy, FL, USA.


U336. Find a closed form for the sum En =∑n

k=0

(n

2k+1

)3k. Deduce the value of constant c such that

0 < limn→∞Encn <∞, and the limit itself.

Proposed by Ángel Plaza, Universidad de Las Palmas de Gran Canaria, Spain

Solution by Haroun Meghaichi, University of Science and Technology Houari Boumediene, Algiers, Alge-riaLet us consider the general case En(x) =

∑nk=0

(n

2k+1

)xk where x > 0. We have

En(x) =1√x

n∑k=0

(n

2k + 1

)(√x)2k+1

=1

2√x

((1 +√x)n

+(1−√x)n)

In this particluar case x = 3 then

En = En(3) =1

2√

3

((1 +√

3)n

+(

1−√

3)n)

Now, it is clear that c = 1 +√

3 (since∣∣1−√3

∣∣ < 1) and the limit is 12√3.

Also solved by Daniel Lasaosa, Pamplona, Spain; Kwonil Kobe Ko, Cushing Academy, MA, USA; Mah-moud Ezzaki and Chakib Belgani; Adnan Ali, A.E.C.S-4, Mumbai, India; Arkady Alt, San Jose, CA,USA; Byeong Yeon Ryu, The Hotchkiss School, Lakeville, CT, USA; David E. Manes, Oneonta, NY, USA;Henry Ricardo, New York Math Circle; Moubinool Omarjee Lycée Henri IV, Paris France; Robert Bosch,Archimedean Academy, FL, USA; Albert Stadler, Herrliberg, Switzerland.


Olympiad problems

O331. Let ABC be a triangle, let ma, mb, mc be the lengths of its medians, and let p0 be the semiperimeterof its orthic triangle. Prove that

1

ma+

1

mb+

1

mc≤ 3√

3

2p0

Proposed by Mircea Lascu, Zalau, Romania

Solution by Robert Bosch, Archimedean Academy, FL, USAWe start with some well-known formulas for orthic triangle. The area is

S0 =abc| cosA cosB cosC|

2R

and the inradiusr0 = 2R| cosA cosB cosC|

thus the semiperimeter has the formula 2p0 =2S

Rwhere S is the area of triangle ABC. Now using that

S =aha2

the original inequality can be rewritten in the following form

a · hama

+ b · hbmb

+ c · hcmc≤ 3√

3R.

But,hama

= sinαA where αA is the angle formed by median and side BC. Consequently, using the Law of

Sines in ABC, the above rewrites as

sinA sinαA + sinB sinαB + sinC sinαC ≤3√

3

2.

The above inequality is true because the left side is less than or equal sinA + sinB + sinC, which is less

than3√

3

2by Jensen’s inequality.

Also solved by Sardor Bozorboyev, Lyceum S.H.Sirojjidinov,Tashkent,Uzbekistan; Neculai Stanciu, GeorgeEmil Palade School, Buzău, Romania and Titu Zvonaru, Comăneşti, Romania; Nicusor Zlota, Traian VuiaTechnical College, Focsani, Romania; Arkady Alt, San Jose, CA, USA; Inequaliter and Equaliter; AN-anduudProblem Solving Group, Ulaanbaatar, Mongolia; Kwonil Kobe Ko, Cushing Academy, MA, USA.


O332. Let ABC be a triangle. Prove that

cos4A

sin4B + sin4C+

cos4B

sin4C + sin4A+

cos4C

sin4A+ sin4B≥ 1

6.

Proposed by An Zhen-ping, Xianyang Normal University, China

Solution by Robert Bosch, Archimedean Academy, FL, USABy the Cauchy-Schwarz inequality we have that∑

cyc

cos4A

sin4B + sin4C≥ (cos2A+ cos2B + cos2C)2

2(sin4A+ sin4B + sin4C).

We next show that(cos2A+ cos2B + cos2C)2

2(sin4A+ sin4B + sin4C)≥ 1

6.

The above rewrites as

3(cos2A+ cos2B + cos2C)2 ≥ sin4A+ sin4B + sin4C,

which by Law of Sines may be transformed into

3(12R2 − (a2 + b2 + c2)

)2 ≥ a4 + b4 + c4,

where a, b, c are the sides of triangle ABC and R is the circumradius. Now the idea is to write the symmetricexpressions a2 + b2 + c2 and a4 + b4 + c4 as functions of s, r, R, with s the semiperimeter and r the inradius.To do the latter, we use Newton’s relations, getting

a2 + b2 + c2 = 2s2 − 8rR− 2r2,

a4 + b4 + c4 = 2s4 − 16s2rR− 12s2r2 + 32r2R2 + 16r3R+ 2r4.

In particular, note that by Blundon’s inequality s2 ≤ 4R2 + 4rR+ 3r2 the following inequalities holds:

a2 + b2 + c2 ≤ 8R2 + 4r2,

a4 + b4 + c4 ≤ 32R4 − 32r3R− 16r4.

Finally, we have to prove that16R4 − 96r2R2 + 64r4 + 32r3R ≥ 0.

Dividing by 16R4 the above inequality can be rewritten as

1− 6( rR

)2+ 4

( rR

)4+ 2

( rR

)3≥ 0.

In other words, writing y =r

R, we have to show that polynomial 4y4 + 2y3 − 6y2 + 1 is nonnegative on the

interval I = (0, 1/2]. This suffices due to Euler’s inequality. But

4y4 + 2y3 − 6y2 + 1 = (2y − 1)(2y3 + 2y2 − 2y − 1).

So take g(y) = 2y3 + 2y2 − 2y − 1 and note that g(y) < 0 on I. Indeed,

g(−2) · g(−1) = −5 < 0,

g(−1) · g(0) = −1 < 0,

g(0.6) · g(1) = −1.048 < 0,

the three real roots are located outside I. Also, g(0) = −1 thus the cubic preserves the sign on I.

Also solved by Arkady Alt, San Jose, CA, USA; Kwonil Kobe Ko, Cushing Academy, MA, USA.


O333. Let ABC be a scalene acute triangle and denote by O, I,H its circumcenter, incenter, and orthocenter,respectively. Prove that if the circumcircle of triangle OIH passes through one of the vertices of triangleABC then it also passes through one other vertex.

Proposed by Josef Tkadlec, Charles University, Czech Republic

Solution by Adnan Ali, A.E.C.S-4, Mumbai, IndiaWe prove the statement not only for acute-angled triangles, but for all scalene triangles. Without loss ofgenerality, assume that ∠CAB > ∠ABC > ∠BCA. It is well-known that O,H are isogonal conjugateswith respect to triangle ABC. So, let the circle pass through A. Thus, we know that AOIH is cyclic.Hence ∠IOH = ∠IAH and ∠IHO = ∠IAO. But O,H being isogonal conjugates w.r.t 4ABC, we have∠IAO = ∠IAH. Thus, ∠IHO = ∠IOH ⇒ IO = IH. Now using the Sine Law for 4IOC and 4IHC andIO = IH, we have

CI

sin∠COI=

IO

sin∠IOC=

IH

sin∠IHC=

CI

sin∠CHI

which implies that ∠COI = ∠CHI or ∠COI+∠CHI = 180◦. Similarly, we have either ∠BOI = ∠BHI or∠BOI + ∠BHI = 180◦. It is quite evident that all AOIH,BOIH,COIH cannot be cyclic quadrilaterals.Thus, if ∠BOI+∠BHI = 180◦, then BOIH is cyclic, and that implies that AHOB is cyclic, implying that∠AOB = ∠AHB ⇒ 2∠ACB = 180◦ − ∠ACB ⇒ ∠ACB = 60◦, which is a contradiction, as the smallestangle of a scalene triangle is always less than 60◦. Hence ∠COI +∠CHI = 180◦ and COIH is cyclic. Thuswe have shown more than what was to be proved, i.e. when the circumcircle of OIH passes through onevertex, it must pass through the other vertex provided that the vertex is opposite to either the largest or theshortest side of the scalene triangle.

Also solved by Daniel Lasaosa, Pamplona, Spain; Mahmoud Ezzaki and Chakib Belgani; Sardor Bo-zorboyev, S.H.Sirojjidinov lyceum, Tashkent, Uzbekistan; Francisco Javier García Capitán, I.E.S ÁlvarezCubero, Priego de Córdoba, Spain.


O334. Let a, b, c be positive real numbers such that a, b, c ≥ 1. Prove that

1

(a3 + 1)2+

1

(b3 + 1)2+

1

(c3 + 1)2≥ 3

2 (a2b2c2 + 1).

Proposed by İlker Can Çiçek, Instanbul, Turkey

Solution by Paolo Perfetti, Università degli studi di Tor Vergata Roma, Roma, ItalyWe will use the known inequality

x, y ≥ 1 =⇒ 1

1 + x+

1

1 + y≥ 2

1 +√xy

(1)

By x2 + y2 ≥ (x+ y)2/2 ≥ 0 we get

1

2

1

(1 + a3)2+

1

2

1

(1 + b3)2≥ 1

4

(1

1 + a3+

1

1 + b3

)2

and by (1)

1

4

(1

1 + a3+

1

1 + b3

)2

≥ 1(1 + (ab)

32

)2We obtain ∑

cyc

1

(1 + a3)2− 3

2(a2b2c2 + 1)≥∑cyc

1

(1 + (ab)32 )2− 3

2(a2b2c2 + 1)

.= F (a, b, c)

The inequality is symmetric so we suppose b ≤ c ≤ a and show that

F (a, b, c) ≥ F (√ab,√ab, c)

namely

1

(1 + (ab)32 )2

+1

(1 + (bc)32 )2

+1

(1 + (ca)32 )2≥ 1

(1 + (ab)32 )2

+2

(1 + (√abc)

32 )2

or

1

(1 + (bc)32 )2

+1

(1 + (ca)32 )2≥ 2

(1 + (√abc)

32 )2

(2)


Moreover

1

(1 + (bc)32 )2

+1

(1 + (ca)32 )2≥ 1

2

(1

(1 + (bc)32 )

+1

(1 + (ca)32 )

)2

and a further use of (1) yields

1

2

(1

(1 + (bc)32 )

+1

(1 + (ca)32 )

)2

≥ 1

2

4(1 +

√(bc)

32 (ca)

32

)2 =2(

1 + c32

√(ab)

32

)2

which is (2).Standard theorems says that

F (a, b, c) ≥ F (a, a, a) = 0

and we are done.

Also solved by Arkady Alt, San Jose, CA, USA; Daniel Lasaosa, Pamplona, Spain; Robert Bosch, Archi-medean Academy, FL, USA; Nicusor Zlota, Traian Vuia Technical College, Focsani, Romania; Adnan Ali,A.E.C.S-4, Mumbai, India; Adithya Bhaskar, Atomic Energy Central School - 2 , Mumbai , India; SardorBozorboyev, S.H.Sirojjidinov lyceum, Tashkent, Uzbekistan; Inequaliter and Equaliter; Mahmoud Ezzaki andChakib Belgani.


O335. Determine all positive integers n such that fn(x, y, z) = x2n + y2n + z2n − xy − yz − zx dividesgn(x, y, z) = (x− y)5n + (y − z)5n + (z − x)5n, as polynomials in x, y, z with integer coefficients.

Proposed by Dorin Andrica, Babeş-Bolyai University, Romania

Solution by Daniel Lasaosa, Pamplona, SpainA necessary condition is that fn(u, v, w) divides gn(u, v, w) for any triple of integers (u, v, w). In particular,taking (x, y, z) = (2, 1, 0), we find that 22n − 1 must divide 2 + (−2)5n.

For even n, we have that 2(25n−1 + 1

)must be a multiple of (2n − 1) (2n + 1). Now,

25n−1 + 1 = (2n + 1)(24n−1 − 23n−1 + 22n−1 − 2n−1

)+ 2n−1 + 1,

and clearly 2n + 1 > 2n−1 + 1 > 0 for any positive integer n, or no even n satisfies the condition given in theproblem statement.

For odd n, we have that 2(25n−1 − 1

)must be a multiple of (2n − 1) (2n + 1). Now,

25n−1 − 1 = (2n + 1)(24n−1 − 23n−1 + 22n−1 − 2n−1

)+ 2n−1 − 1,

and for any n ≥ 3, we have 2n + 1 > 2n−1 − 1 > 0, or only n = 1 may satisfy the condition given in theproblem statement.

For n = 1, we have

g1(x, y, z) =(5x2(z − y) + 5y2(x− z) + 5z2(y − x)

)f1(x),

and we conclude that n = 1 is the only solution to this problem.

Also solved by Albert Stadler, Herrliberg, Switzerland


O336. Let a, b, c be positive distinct real numbers and let u, v, w be positive real numbers such that

a+ b+ c = u+ v + w and (a2 − bc)r + (b2 − ac)s+ (c2 − ab)t ≥ 0

for (r, s, t) = (u, v, w), (r, s, t) = (v, w, u), (r, s, t) = (w, u, v). Prove that

xaybzc + xbycza + xcyazb ≥ xuyvzw + xwyuzv + xvywzu

for all nonnegative numbers x, y, z.

Proposed by Albert Stadler, Switzerland

Solution by Arkady Alt , San Jose ,California, USABy the weighted AM-GM inequality,

pxaybzc + qxbycza + rxcyazb ≥ (p+ q + r)(xpa+qb+rcypb+qc+razpc+qa+rb

) 1p+q+r

,

for any positive reals p, q, r.

We now search for p, q, r such thatpa+ qb+ rc = upb+ qc+ ra = vpc+ qa+ rb = w

or in matrix form

a b cb c ac a b

pqr

=

uvw

.

Since a+ b+ c = u+ v + w then adding all equations we obtain

(p+ q + r) (a+ b+ c) = u+ v + w ⇐⇒ p+ q + r = 1.

Since det

a b cb c ac a b

=∑cyca(bc− a2

)= −

∑cyca(a2 − bc

)= − (a+ b+ c)

(a2 + b2 + c2

)

and det

u b cv c aw a b

=∑cycu(bc− a2

)= −

∑cycu(a2 − bc

), then p =

∑cycu(a2 − bc

)(a+ b+ c) (a2 + b2 + c2)

,

and cyclically

q =

∑cycv(a2 − bc

)(a+ b+ c) (a2 + b2 + c2)

, r =

∑cycw(a2 − bc

)(a+ b+ c) (a2 + b2 + c2)

,

where, we get p, q, r all positive due condition

(a2 − bc)r + (b2 − ac)s+ (c2 − ab)t ≥ 0

for (r, s, t) = (u, v, w), (r, s, t) = (v, w, u), (r, s, t) = (w, u, v).For the obtained p, q, rwe have

pxaybzc + qxbycza + rxcyazb ≥ xuyvzw,

and similarly

qxaybzc + rxbycza + pxcyazb ≥ xwyuzv, rxaybzc + pxbycza + qxcyazb ≥ xvywzu.

Adding these inequalities and taking in account that p+ q + r = 1 we finally obtain

xaybzc + xbycza + xcyazb ≥ xuyvzw + xwyuzv + xvywzu.


mr 2 2015 solutions

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