solutions - canadian mathematical society94/ solutions 4314.proposed by michel bataille. let nbe a...

90/ Solutions

SOLUTIONSNo problem is ever permanently closed. The editor is always pleased to consider forpublication new solutions or new insights on past problems.

Statements of the problems in this section originally appear in 2018: 44(2), p. 69–72.

4300?. Proposed by Leonard Giugiuc.

Let a, b and c be positive real numbers with a + b + c = ab + bc + ca > 0. Proveor disprove that

√24ab+ 25 +

√24bc+ 25 +

√24ca+ 25 ≥ 21.

Arkady Alt pointed out some defects in the solution of 4300 published in 43(10).We present a complete solution here.

We received 3 submissions of which 2 were correct and complete. We feature thesolution by the proposer modified by the editor.

Let a + b + c = ab + bc + ac = k. Since (a + b + c)2 ≥ 3(ab + bc + ac), we havek2 ≥ 3k, and since k > 0 it follows that k ≥ 3. Assume, without loss of generality,that bc = max{ab, bc, ac}. Then bc ≥ 1, and a = b+c−bc

b+c−1 . Setting b + c = 2s and

bc = p2, we know that s ≥ p, p ≥ 1, and a = 2s−p22s−1 . We consider two cases.

Case 1: 1 ≤ p ≤ 2. For any such p, let fp(t) = 2s−p22t−1 on [p,∞). Since p2 ≥ 1, we

know that fp is increasing, which implies that a = 2s−p22s−1 ≥

2p−p22p−1 . If we apply the

AM-GM inequality we have:

√24ab+ 25 +

√24ac+ 25 ≥ 2((24ab+ 25)(24ac+ 25))1/4.

Then

2((24ab+ 25)(24ac+ 25))1/4

= 2

Ç576

Å2s− p2

2s− 1

ã2p2 + 1200

Å2s− p2

2s− 1

ãs+ 625

å1/4

≥ 2

Ç576

Å2p− p2

2p− 1

ã2p2 + 1200

Å2p− p2

2p− 1

ãp+ 625

å1/4

= 2

Çï24

Å2− p2p− 1

ãp2 + 25

ò2å1/4

= 2

24

Å2− p2p− 1

ãp2 + 25.

Crux Mathematicorum, Vol. 45(2), February 2019

Solutions /91

So we have

√24ab+ 25 +

√24ac+ 25 ≥ 2

24

Å2− p2p− 1

ãp2 + 25,

and√

24bc+ 25 =√

24p2 + 25. So it suffices to show that

√24p2 + 25 + 2

24

Å2− p2p− 1

ãp2 + 25 ≥ 21.

Since 1 ≤ p ≤ 2, it follows that√24p2 + 25 + 2

24

Å2− p2p− 1

ãp2 + 25 ≥

√24p2 + 25 + 2

√73− 24p

Using basic calculus, it follows that the function f : [1, 2] −→ R,

f(p) =√

24p2 + 25 + 2√

73− 24p

is strictly decreasing on [1, 5/3] and it is strictly increasing on [5/3, 2]. Hence ithas a minimum at 5/3 and the minimum value is more than 21.

Case 2: p > 2. Then we have√

24ab+ 25 +√

24ac+ 25 +√

24bc+ 25 ≥ 10 +√

24p2 + 25 > 10 + 11 = 21.

4311. Proposed by Mihaela Berindeanu.

Let A and B be two matrices in M3 (Z) with AB = BA and detA = detB = 1.Find the possible values for det

(A2 +B2

)knowing that

det(A2 + 2AB + 4B2

)− det

(A2 − 2AB + 4B2

)= −4.

We received 4 correct solutions and one incomplete submission. We will feature asolution by Leonard Guigiuc.

Since det(B) = 1, then det(B−1) = 1 and B−1 ∈ M3(Z). This implies thatdet(AB−1) = 1 and AB−1 ∈M3(Z). Since AB = BA, we set C = AB−1 to get

det(A2+2AB+4B2)−det(A2−2AB+4B2) = det(C2+2C+4I3)−det(C2−2C+4I3),

and det(A2 +B2) = det(C2 + I3).

Consider the polynomial f(x) = det(C−xI3) over C. Then f(x) = 1−kx+tx2−x3for all x ∈ C with k and t integers.

Let u = 1/2(−1 + i√

3). Then C2 + 2C + 4I3 = (C − 2uI3)(C − 2u2I3), whichimplies

det(C2 + 2C + 4I3) = det(C − 2uI3) · det(C − 2u2I3)

= f(2u) · f(2u2)

= 49 + 4k2 + 16t2 − 14k + 28t+ 8kt.

Copyright © Canadian Mathematical Society, 2019

92/ Solutions

Similarly,

det(C2 − 2C + 4I3) = 81 + 4k2 + 16t2 − 18k − 36t− 8kt.

Thusdet(C2 + 2C + 4I3)− det(C2 − 2C + 4I3) = −4

if and only if k + 16t + 4kt = 7 or, equivalently, 4t(k + 4) = 7 − k. Since clearly,k 6= −4, we have 4t = −1 + 11/(k + 4), which implies 11/(k + 4) ∈ Z. Hencek + 4 ∈ {±1,±11}. Since t is an integer as well, we obtain k = −5 and t = 3 ork = 7 and t = 0. Observe that

det(C2 + I3) = det(C − I3) · det(C + I3) = f(i) · f(−i) = (k − 1)2 + (t− 1)2.

In conclusion, det(A2 +B2) ∈ {37, 52}.

4312. Proposed by William Bell.

Prove that∞∑r=1

1

2rtanh

( x2r

)= cothx− 1

x.

Nine correct solutions were received. Six followed the strategy of the first solutionand three the strategy of the second.

Solution 1.

Since tanhu = 2 coth 2u− cothu,

n∑r=1

1

2rtanh

x

2r=

n∑r=1

Å1

2r−1coth

x

2r−1− 1

2rcoth

x

2r

ã= cothx− 1

2ncoth

x

2n.

Since, by l’Hopital’s Rule, for example, limv→0 v coth v = 1,

∞∑r=1

1

2rtanh

x

2r= cothx− 1

xlimn→∞

x

2ncoth

x

2n= cothx− 1

x.

Solution 2.

For n ≥ 1, consider the product

Pn(x) = cosh(x

2

)cosh

(x4

). . . cosh

( x2n

).

Since

Pn(x) sinh( x

2n

)=

1

2Pn−1(x) sinh

( x

2n−1

),

an induction argument leads to

Pn(x) =sinhx

2n sinh(x2n

) .Crux Mathematicorum, Vol. 45(2), February 2019

Solutions /93

With D denoting differentiation, we have that

n∑r=1

1

2rtanh

x

2r= D

(n∑r=1

ln coshx

2r

)= D ln

(n∏r=1

coshx

2r

)= D lnPn(x)−D(ln sinhx− n ln 2− ln sinh

x

2n)

= cothx− 1

2ncoth

x

2n,

Let n→∞ to obtain

∞∑r=1

1

2rtanh

( x2r

)= cothx− 1

x.

4313. Proposed by Marian Cucoanes and Leonard Giugiuc.

Let I be the incenter of triangle ABC, and denote by Ha, Hb and Hc the ortho-centers of triangles IBC, ICA and IAB, respectively. Prove that triangles ABCand HaHbHc have the same area.

We received six submissions, all correct, and feature the solution by MohammedAassila.

The result holds for any point P in the plane of triangle ABC that is not on a linejoining two of its vertices; the argument is therefore not restricted to the specialcase of the problem, namely P = I. It is based on a familiar property of the mixedproduct [−→p ,−→q ], which for vectors −→p = 〈p1, p2〉 and −→q = 〈q1, q2〉 in R2 is just thedeterminant,

[−→p ,−→q ] =

∣∣∣∣ p1 p2q1 q2

∣∣∣∣ = p1q2 − p2q1.

Specifically, given a pair of triangles ABC and A′B′C ′ in the plane, we have

[−−→BA′,

−−→AB′] + [

−−→CB′,

−−→BC ′] + [

−−→AC ′,

−−→CA′] = [

−−−→A′B′,

−−→A′C ′]− [

−−→AB,

−→AC].

This identity can be found as a straightforward exercise in texts that deal with themixed product. Of course, when A′, B′, and C ′ are, respectively, the orthocentresof triangles PBC,PCA, and PAB, then each of the quantities on the left are zero

(because the vectors−−→BA′ and

−−→AB′ are both perpendicular to the line PC, etc.).

We are therefore left with

[−−−→A′B′,

−−→A′C ′] = [

−−→AB,

−→AC],

which says (under the assumption that A′, B′, C ′ are the appropriate orthocentres)that the areas of triangles ABC and A′B′C ′ are equal.


94/ Solutions

4314. Proposed by Michel Bataille.

Let n be a positive integer. Evaluate in closed form

n∑k=1

k2k ·(nk

)(2n−1k

) .We received three solutions, and we present two of them.

Solution 1, by Paolo Perfetti, slightly edited.

Denote the kth summand by sk; that is,

sk = k2k(nk

)(2n−1k

) = k2k(nk

)k!(2n− k − 1)!

(2n− 1)!.

We will use the beta function, β(x, y) =∫ 1

0tx−1(1 − t)y−1 dt defined for com-

plex x, y such that Re(x),Re(y) > 0. One of the well known properties of thebeta function (which follows from its relationship to the gamma function) is that

β(x, y) = (x−1)!(y−1)!(x+y−1)! for x, y positive integers. Hence we can write

sk = k · 2k ·Çn

k

å· 2n · β(k + 1, 2n− k),

and we haven∑k=1

sk =n∑k=1

Çk2kÇn

k

å· 2n

∫ 1

0

tk(1− t)2n−k−1 dtå

= 2n

∫ 1

0

(1− t)2n−1n∑k=1

k

Çn

k

åÅ2t

1− t

ãkdt. (1)

Note that for any y we have the equalityn∑k=1

k(nk

)yk = ny(y + 1)n−1:

n∑k=1

k

Çn

k

åyk = n

n∑k=1

Çn− 1

k − 1

åyk = ny

n−1∑k=0

Çn− 1

k

åyk = ny(1 + y)n−1,

with the last equality following from the binomial theorem. Thus, continuing from(1) we get

n∑k=1

sk = 2n

∫ 1

0

(1− t)2n−1Å

2nt

1− t

ãÅ2t

1− t+ 1

ãn−1dt

= 2n2∫ 1

0

(1− t)n−1(2t)(1 + t)n−1 dt

= 2n2∫ 1

0

2t(1− t2)n−1 dt

= 2n2 ·Å− 1

n

ã· (1− t2)n

∣∣∣∣10

= 2n.


Solutions /95

Solution 2, by the proposer.

Denote the kth summand by sk as in the previous solution.

Let

uk = −2k(nk

)(2nk

) = −2kn!(2n− k)!

(2n)!(n− k)!

for k = 1, 2, . . . , n and un+1 = 0. Note that for k = 1, 2, . . . , n− 1 we have

uk+1 − uk = 2k · n!

(2n)!· (2n− k − 1)!

(n− k − 1)!·Å−2 +

2n− kn− k

ã= 2k · n!

(2n)!· (2n− k − 1)!

(n− k − 1)!· k

n− k

= k · 2k · n!

(2n)!· (2n− k − 1)!

(n− k)!=sk2n.

The equality uk+1 − uk = sk2n holds trivially for k = n since un+1 = 0.

It follows that

n∑k=1

sk = 2n

(n∑k=1

uk+1 − uk

)= 2n(un+1 − u1) = 2n.

4315. Proposed by Moshe Stupel, modified by the editors.

Let H be the orthocenter of triangle ABC, and denote by R, r, and r′ respectivelythe circumradius, inradius, and radius of the excircle that is opposite vertex A.Prove that HA+ r′ = 2R+ r.

We received 11 submissions, all substantially correct. We present a combinationof the solutions from Leonard Giugiuc and Cristobal Sanchez-Rubio.

The statement of the problem is not quite correct. The identity we shall prove is

2R cosA+ r′ = 2R+ r. (1)

Editor’s comments. Giugiuc and just one other solver (Pranesachar) stated ex-plicitly that HA = 2R cosA when 0◦ < ∠A ≤ 90◦; otherwise (when ∠A is obtuse)HA = −2R cosA. All other submissions tacitly took HA to be a directed lengthby setting it equal to 2R cosA. If you prefer to assume lengths to be nonnegative,then the correct identity in terms of HA is r′ ±HA = 2R + r, with the negativesign used when ∠A is obtuse.

We shall make use of familiar formulas for the circumradius

R =abc

4sr, (2)


96/ Solutions

and for the area,

rs = r′(s− a) =»s(s− a)(s− b)(s− c). (3)

Using expressions for R and r′ (from (2) and (3)) reduces identity (1) (in the form2R(cosA− 1) = r − r′) to

2abc

4sr(cosA− 1) = r

Å1− s

s− a

ã= − ra

s− a,

so that the identity to be established is equivalent to

bc (1− cosA)

2sr=

r

s− a. (4)

But because

1− cosA = 1 +a2 − b2 − c2

2bc

=a2 − (b− c)2

2bc

=(a+ b− c) (a− b+ c)

2bc

=2 (s− c) · 2 (s− b)

2bc

=2 (s− b) (s− c)

bc,

equation (4) reduces to

(s− a) (s− b) (s− c) = sr2,

which is the square of Heron’s formula (see (3) above).

4316. Proposed by Daniel Sitaru.

Let f : [0, 11]→ R be an integrable and convex function. Prove that∫ 5

3

f(x)dx+

∫ 8

6

f(x)dx ≤∫ 2

0

f(x)dx+

∫ 11

9

f(x)dx.

Ten correct solutions were received. Most followed the procedure of Solution 2.

Solution 1, by Roy Barbara.

Let g(x) = ax+b be the linear function that satisfies g(3) = f(3) and g(8) = f(8).

Because f(x) is convex, f(x) ≥ g(x) when 0 ≤ x ≤ 3 or 8 ≤ x ≤ 10, andf(x) ≤ g(x) when 3 ≤ x ≤ 8. The left side does not exceed∫ 5

3

g(x) dx+

∫ 8

6

g(x) dx = 22a+ 4b,


Solutions /97

and the right side is not less than∫ 2

0

g(x) dx+

∫ 11

9

g(x) dx = 22a+ 4b.

The result follows.

Solution 2.

Since f(x) is convex,

f(3 + x) ≤ 2

3f(x) +

1

3f(9 + x) and f(6 + x) ≤ 1

3f(x) +

2

3f(9 + x).

Therefore, ∫ 5

3

f(x) dx+

∫ 8

6

f(x) dx =

∫ 2

0

[f(3 + x) + f(6 + x)] dx

≤∫ 2

0

[f(x) + f(9 + x)] dx

=

∫ 2

0

f(x) dx+

∫ 11

9

f(x)dx.

Solution 3, by Oliver Geupel.

Recall the Hermite-Hadamard Inequality for convex functions:

(b− a)f

Åa+ b

2

ã≤∫ b

a

f(x) dx ≤ 1

2(b− a)(f(a) + f(b).

Therefore ∫ 5

3

f(x) dx ≤ f(3) + f(5)

≤Å

7

9f(1) +

2

9f(10)

ã+

Å5

9f(1) +

4

9f(10)

ã=

4

3f(1) +

2

3f(10),

and, similarly, ∫ 8

6

f(x) dx ≤ f(6) + f(8) ≤ 2

3f(1) +

4

3f(10).

Therefore∫ 5

3

f(x) dx+

∫ 8

6

f(x) dx ≤ 2f(1) + 2f(10) ≤∫ 2

0

f(x) dx+

∫ 11

9

f(x) dx.


98/ Solutions

4317. Proposed by Leonard Giugiuc.

Solve the following system of equations over reals:a+ b+ c+ d = 4,

abc+ abd+ acd+ bcd = 2,

abcd = − 14 .

We received nine correct submissions. We present the solution by Oliver Geupel,modified and expanded by the editor.

Let p = a+ b, q = ab, r = c+ d, s = cd. Then q 6= 0 and s = − 1

4q.

Since

2 = ab(c+ d) + cd(a+ b) = qr + ps = q(4− p)− p

4q,

we have

8q = 4q2(4− p)− p = 16q2 − (4q2 + 1)p

so

p =8q(2q − 1)

4q2 + 1and r = 4− p =

4(2q + 1)

4q2 + 1.

Since the quadratic x2 − px + q has two real roots a and b, its discriminant isnon-negative.

By labourious computations, we find that

p2 − 4q =64q2(2q − 1)2

(4q2 + 1)2− 4q =

4q

(4q2 + 1)2D

where

D = 16q(4q2 − 4q + 1)− (16q4 + 8q2 + 1)

= −16q4 + 64q3 − 72q2 + 16q − 1

= −16(q4 − 4q3 + 92q

2 − q + 116 )

= −16(q2 − 2q + 14 )2

= −16((q − 1 +√32 )(q − 1−

√32 ))2.

Since p2 − 4q2 ≥ 0 and q 6= 0, it follows that

either q < 0 or q = 1±√

3

2(1)

Similarly, since the quadratic x2−rx+s has two real roots c and d, its discriminantis non-negative.


Solutions /99

By labourious computations, we find that

r2 − 4s2 =16(2q + 1)2

(4q2 + 1)2+

1

q

=16q(2q + 1)2 + (4q2 + 1)2

q(4q2 + 1)2

=16

q·q4 + 4q3 + 9

2q2 + q + 1

16

(4q2 + 1)2

=16

q

Åq2 + 2q + 1

4

4q2 + 1

ã2=

16

q

Å(q + 1 +

√32 )(q + 1−

√32 )

4q2 + 1

ã2.

Since r2 − 4s2 ≥ 0 it follows that

either q > 0 or q = −1±√

3

2. (2)

From (1) and (2) it is readily seen that there are four possible values of q given by

q = ±1±√32 .

For the first case q = 1 +√32 , we have

p =8q(2q − 1)

4q2 + 1= 1 +

√3,

r = 4− p = 3−√

3,

s =−1

4q=

−1

4 + 2√

3= −1 +

√3

2.

Solving the equations

a+ b = p, ab = q, c+ d = r, cd = s,

we find by tedious calculations that

(a, b, c, d) =

Å1 +√

3

2,

1 +√

3

2,

1 +√

3

2,

5− 3√

3

2

ã. (3)

Similarly, the second case q = 1−√32 yields

p = 1−√

3, r = 3 +√

3, s = 1−√32 ,

which would lead to the solution

(a, b, c, d) =

Å1−√

3

2,

1−√

3

2,

1−√

3

2,

5 + 3√

3

2

ã. (4)


100/ Solutions

The third case q = −1+√32 is symmetric to the first case with p and q interchanged

with r and s, respectively, leading to a permutation of the solution in (3). Finally,

the fourth case q = −1−√33 would eventually lead to a permutation of the solution

in (4).

In conclusion, all the solutions are given by the two ordered quadruples togetherwith all of their permutations.

4318. Proposed by Thanos Kalogerakis.

Given a pair of intersecting circles (just their circumferences, not their centres), letAB be the common diameter with one end on each circle and neither end insideeither circle. Show how to construct the midpoint of AB using only a straightedgeand prove that your construction is correct.

We received four submissions, but one was incomplete. We present the solutionsent in by the Missouri State University Problem-Solving Group, modified by theeditor.

Denote the circle containing A by α, the circle containing B by β, one of the pointswhere α and β intersect by X, and the point other than B where AB meets β byB′. Our preliminary step is to construct the center of α. We note that once wehave the center of α, by the Poncelet-Steiner Theorem we can construct any pointconstructible with compass and straightedge using straightedge alone.


Solutions /101

Since ∠B′XB is inscribed in a semi-circle, XB′ is perpendicular to XB. If XB isnot tangent to α, it meets α in another point Y . Let Z denote the second pointwhere XB′ meets α. Since ∠Y XZ is a right angle, Y Z is a diameter of α and theintersection of Y Z and AB gives the center of α, which we denote by O. If XB istangent to α, then XB′ contains a diameter of α so that B′ = O.

We now construct the midpoint M of AB given that O is the midpoint of thediameter AA′. Choose P to be a point between A and Y on the segment AY , anddefine

Q = PA′ ∩ Y O, R = QA ∩ Y A′, and S = Y O ∩ PR,

as in the figure on the left. It is a known property of the quadrangle AA′RPthat PR is parallel to AA′ and, therefore, that S is the midpoint of PR. [For aproof featuring harmonic conjugates see the Crux article “Problem Solver’s ToolkitNo. 5: Harmonic Sets Part 2”, 2013: 174-177, especially Ex. 3, page 176.]

Finally, define C = BR ∩ AP and M = CS ∩ AB, as in the figure on the right.Because S is the midpoint of the segment PR, M must be the midpoint of theparallel segment AB, as required.

Comment. The problem as stated dealt with intersecting circles. We can stillconstruct the center of α by straightedge when the given circles α and β aretangent (so that X,A′, and B′ coincide in a single point). Choose any point Pdifferent from A and X on α such that PB intersects β in a second point Q, and αin a second point S. Denote the second point where XQ meets α by R. Now QRis perpendicular to PB since ∠XQB is inscribed in a semicircle. Similarly, QRis perpendicular to AR since ∠ARX is inscribed in a semicircle. Therefore PBis parallel to AR. The convex quadrilateral with vertices A,P, S,R is an isoscelestrapezoid. Let T be the intersection of PR and AS and let U be the intersectionof AP and RS. Then TU contains a diameter of α, so the intersection of TU andAB gives the center of α.

Editor’s comments. The featured solution requires drawing 10 lines, but the in-complete submission found M drawing only 5 lines. Unfortunately that solutioncame as a diagram with no accompanying justification. Perhaps the author justwanted to challenge his fellow readers, so we pass along that challenge as one ofthis month’s problems, namely number 4414.


102/ Solutions

4319. Proposed by Marius Dragan.

Let x1, x2, . . . , xn ∈ (0,+∞), n ≥ 2, α ≥ 32 such that xα1 + xα2 + · · · + xαn = n.

Prove the following inequality:

n∏i=1

(1 + xi + xα+1i ) ≤ 3n.

The original proposal contained a typo corrected here.

We received 1 correct solution, by the proposer, and we present it here.

Setting a =1

α∈ï0,

2

3

ò, we consider the function f : (0,∞)→ R defined by

f(x) = ln(1 + xa + xa+1).

We have

f ′(x) =axa−1 + (a+ 1)xa

1 + xa + xa+1,

f ′′(x) =−(a+ 1)x2a− 2ax2a−1− ax2a−2 + (a2+ a)xa−1 + (a2− a)xa+2

(1 + xa + xa+1)2.

We show that f ′′(x) < 0, for all x ∈ (0,∞). It will be sufficient to prove that forall x ∈ (0,∞),

−(a+ 1)xa+2 − 2axa+1 − axa + (a2 + a)x+ a2 − a ≤ 0

or(a2 + a)x < a− a2 + axa + 2axa+1 + (a+ 1)xa+2.

For this it will be enough to prove that

(a2 + a)x < a− a2 + 2axa+1

or

x <1− aa+ 1

+2

a+ 1xa+1.

From the power mean inequality, we obtain

a

1 + a· 1− a

a+

1

1 + a· 2xa+1 ≥

Å1− aa

ã aa+1

(2xa+1)1

1+a =

Å1− aa

ã aa+1

21

1+ax.

It will thus be sufficient to prove that

Åa

1− a

ãa≤ 2. But since a ∈

ï0,

2

3

ò, we

have 1 ≤2

1− a≤ 2 or

Åa

1− a

ãa≤ 2a. But since 2a < 2, the claimed inequality

holds. Thus f is a concave function.


Solutions /103

From Jensen’s inequality we have

n∑i=1

f(xαi ) ≤ nf

Ün∑i=1

xαi

n

êor

n∑i=1

f(xαi ) ≤ nf(1) or lnn∏i=1

(1 + xi + xα+1i ) ≤ n ln 3.

4320. Proposed by Abhay Chandra.

For positive real numbers a, b, c, d, prove that

(a+ b)(a+ c)(a+ d)(b+ c)(b+ d)(c+ d) ≥ 16 (a+ b+ c+ d)4√a5b5c5d5.

We received 8 submissions, all correct. We present the proof by SefketArslanagic.

We first establish the following lemma.

Lemma If a, b, c, d are positive reals, then

(a+ b)(b+ c)(c+ d)(d+ a) ≥ (a+ b+ c+ d)(abc+ bcd+ cda+ dab).

Proof We have (a+ b)(b+ c)(c+ d)(d+ a)

= (ac+ bd+ ad+ bc)(ac+ bd+ ab+ cd)

= (ac+ bd)2 +∑cyc

a2(bc+ cd+ db)

≥ 4abcd+∑cyc

a2(bc+ cd+ db)

= (a+ b+ c+ d)(abc+ bcd+ cda+ dab), as claimed.

By the lemma and the AM-GM Inequality, we have

(a+ b)(a+ c)(a+ d)(b+ c)(b+ d)(c+ d)

≥ (a+ b+ c+ d)(abc+ bcd+ cda+ dab) · 2√ac · 2

√bd

≥ (a+ b+ c+ d)44√a3b3c3d3 · 4

√abcd

= 16(a+ b+ c+ d)4√a5b5c5d5, completing the proof.

Equality holds if and only if a = b = c = d.


solutions - canadian mathematical society94/ solutions 4314.proposed by michel bataille. let nbe a...

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