Solutions of J´ ozsef Wildt International Mathematical Competition

Vol. 17, No.2, October 2009, pp 773-796 ISSN 1222-5657, ISBN 978-973-88255-5-0, www.hetfalu.ro/octogon

773

Solutions of J´ ozsef Wildt International Mathematical Competition

The Edition XIX^th, 2009

Mih´aly Bencze and Zhao Chang Jian³⁷

W1. Consider the functionf : (0,1)→R defined by f(x) = 1

xln

µ1 +x 1−x

¶

Since f(x) = 2P^∞

k=0 x^2k

2k+1 for|x|<1, thenf⁰(x) = 4P^∞

k=0 kx^2k−1

2k+1 (|x|<1) and f⁰⁰(x) = 4P^∞

k=0

k(2k−1)x^2k−2

2k+1 (|x|<1).. Therefore, f⁰(x)>0 and f⁰⁰(x)>0 for allx∈(0,1), andf is increasing and convex.

Applying Jensen‘s inequality, we havef¡_a+b+c

3

¢≤ ^{f(a)+f(b)+f(c)}₃ or equivalently

3 a+b+cln

µ3 + (a+b+c) 3−(a+b+c)

¶

≤

≤ 1 3

"

ln

µ1 +a 1−a

¶_1/a + ln

µ1 +b 1−b

¶_1/b + ln

µ1 +c 1−c

¶_1/c#

Taking into account thata+b+c= 1 and the properties of logarithms, we get

3

sµ1 +a 1−a

¶_1/a ln

µ1 +b 1−b

¶_1/b ln

µ1 +c 1−c

¶_1/c

≥8 (1)

37Received: 28.09.2009

2000Mathematics Subject Classification. 11-06 Key words and phrases. Contest

WLOG we can assume thata≥b≥c. We have, ¹_a ≤ ¹_b ≤ ¹_c and g(a)≥g(b)≥g(c),whereg is the increasing function defined by g(x) = ln

³1+x 1−x

´

.Applying rearragement‘s inequality, we get 1

bg(a) +1

cg(b) +1

ag(c)≥ 1

ag(a) +1

bg(b) +1

cg(c) or µ1 +a

1−a

¶_1/bµ 1 +b 1−b

¶_1/cµ 1 +c 1−c

¶_1/a

≥

µ1 +a 1−a

¶_1/aµ 1 +b 1−b

¶_1/bµ 1 +c 1−c

¶_1/c

From the preceeding and (1) we obtain

3

sµ1 +a b+c

¶_1/aµ 1 +b c+a

¶_1/bµ 1 +c a+b

¶_1/c

=

= ³

sµ1 +a 1−a

¶_1/bµ 1 +b 1−b

¶_1/cµ 1 +c 1−c

¶_1/a

≥ ³

sµ1 +a 1−a

¶_1/aµ 1 +b 1−b

¶_1/bµ 1 +c 1−c

¶_1/c

≥8 Likewise, applying rearrangement‘s inequality again, we get

1

cg(a) +1

ag(b) +1

bg(c)≥ 1

ag(a) + 1

bg(b) +1

cg(c) and

3

sµ1 +a b+c

¶_1/cµ 1 +b c+a

¶_1/aµ 1 +c a+b

¶_1/b

= ³

sµ1 +a 1−a

¶_1/cµ 1 +b 1−b

¶_1/aµ 1 +c 1−c

¶_1/b

≥

≥ ³

sµ1 +a 1−a

¶_1/aµ 1 +b 1−b

¶_1/bµ 1 +c 1−c

¶_1/c

≥8 Multiplying up the preceeding inequalities yields,

3

sµ 1 +a b+c

¶¹

b+¹_c µ 1 +b c+a

¶¹

c+¹_aµ 1 +c a+b

¶¹

a+¹_b

≥64

from which the statement follows. Equality holds whena=b=c= 1/3, and we are done.

W2. The function f(x) can be written in the form

f(x) =

¯¯

¯¯

¯¯

¯¯

¯

1 1 1 1

lnx lnx² lnx³ lnx⁴ (lnx)² ¡

lnx²¢₂ ¡

lnx³¢₂ ¡ lnx⁴¢₂ (lnx)³ ¡

lnx²¢₃ ¡

lnx³¢₃ ¡ lnx³¢₃

¯¯

¯¯

¯¯

¯¯

¯ Developing the Vandermonde‘s determinant, we get

f(x) = (lnx) 2 (lnx) 3 (lnx) (lnx) 2 (lnx) (lnx) = 12 (lnx)⁶ LetI_n=R

(lnx)ⁿdx, (n≥1).We have to computeI₆.We will argue integrating by parts. Let

u= (lnx)⁶ du= ⁶_x(lnx)⁵dx v=x dv=dx

¾

Then,

I₆=x(lnx)⁶−6 Z

(lnx)⁵dx=x(lnx)⁶−6I₅

Likewise,I₅=x(lnx)⁵−5I₄, I₄=x(lnx)⁴−4I₃, I₃ =x(lnx)³−3I₂, I₂ =x(lnx)²−2I₁ andI₁=xlnx−x+K.Therefore,

Area(A) = Ze

1

f(x)dx= 12 Ze

1

(lnx)⁶dx=

= h

12x(lnx)⁶−72x(lnx)⁵+ 360x(lnx)⁴−1440x(lnx)³+ 4320x(lnx)²

−8640 (lnx) + 8640x]^e₁ '4.13 and we are done.

W3. We have ϕ(n)

n =Y

p|n

µ 1− 1

p

¶

and Ψ (n)

n =Y

p|n

µ 1 +1

p

¶ ,

wherep are the prime divisors ofn. This implies a well-known property, namely that forn≥3,ϕ(n) and Ψ (n) are even.

1). Let now nbe odd;n≥3.

Then asϕ(n) is even andn+ Ψ (n) odd;ϕ(n) dividesn+ Ψ (n) is impossible.

Forn= 1 the property is true, asϕ(1) = 1.Therefore,n= 1 is the single odd solution to the problem.

2). Let nbe even, and putn= 2^kN,where N is odd. Then

ϕ(n) = 2^k−1·ϕ(N) and n+Ψ (n) = 2^kN+2^k−1·3·Ψ (N) = 2^k−1[2N + 3Ψ (N)]

Thus we should have also:

ϕ(N)|[2N+ 3Ψ (N)] (∗)

IfN = 1, then (*) holds true. Letn >1,N = Q^r

i=1

p^a_iⁱ (prime representation).

Then from (*) we get

Yr

i=1

p^a_iⁱ⁻¹(p₁−1) divides

Yr

i=1

p^a_iⁱ⁻¹

"

2 Yr

i=1

p_i+ 3 Yr

i=1

(p_i+ 1)

# , which means that

Yr

i=1

(p_i−1)|

"

2 Yr

i=1

p_i+ 3 Yr

i=1

(p_i+ 1)

#

(∗∗) (thus, it is sufficient that in (*),N to be squarefree). Let r≥2. Then the left side of (**) is divisible by 4, but the left side is not (divisible only by 2).

Thusr = 1, which gives, withp₁ =p:

p−1|[2p+ 3 (p+ 1)] = 5 (p−1) + 8

Therefore, p−1|8. This is possible only forp= 3 and 5 implying that N = 3^a orN = 5^a.

This means that n= 2^k1 or 2^k3^a or 2^k5^a.

In conclusion, all solutions to the problem aren= 1 orn= 2^k,orn= 2^k3^a, orn= 2^k·5^a,wherek≥1, a≥1 are arbitrary positive integers.

W4. i). It is well-known that a|b⇒ ^ϕ(a)_a ≤ ^ϕ(b)_b .Letp be a prime. Since 2^p+ 1 is always multiple of 3, we get

ϕ(2^p+ 1)

2^p+ 1 ≥ ϕ(3) 3 = 2

3

Thusϕ(2^p+ 1)≥ ²₃(2^p+ 1)>2^p−1,as 2·2^p+ 2>3·2^p−1 or equivalently 4·2^p−1+ 2>3·2^p−1.Thus for m=p, the second assertion follows.

ii). Let p₁< p₂ < ... < p_n< ...be all primes p≡3 (mod 8). Note that for such primes p we have

p|2^p−1−1 =

³

2^p−12 −1

´ ³

2^p−12 + 1

´ ,

and p divides the second paranthesis and not the first (indeed, ifp divides the first, then since ^p−1₂ is odd, we would get that 2 is a quadratic residue mod p, false forp≡3 (mod 8) [see textbooks of Number Theory]).

Thusp divides 2^p−12 + 1.

Now let nbe the smallest number such that µ

1− 1 p₁

¶ µ 1− 1

p₂

¶ ...

µ 1− 1

p_n

¶

< 1 3 (this is true, since the left side tends to 0 asn→ ∞).

Put now

k=l.c.m.

·p₁−1

2 , ...,p_n−1 2

¸

This numberkis odd as has the property that p₁p₂...p_n|2^k+ 1, so

ϕ¡

2^k+ 1¢

2^k+ 1 ≤ ϕ(p₁...p_n) p₁...p_n =

µ 1− 1

p₁

¶ ...

µ 1− 1

p_n

¶

< 1 3 Thusϕ¡

2^k+ 1¢

< ^2k₂⁺¹ <2^k−1 fork≥k₀. This proves the first assertion of the problem.

W5. From the given assumption the case whenp₁= 2 and p₂ = 2 is not considered. Thus we let p₂ = 2x+ 1 forx∈N.Let us assume that the given equation accept at least one integer solution.

Then the equation

µp₂−1 2

¶_p1 +

µp₂−1 2

¶_p1

=aⁿ is written in the form

x^p1 + (x+ 1)^p1 =aⁿ (1) Since p₁ is an odd integer, it follows that

aⁿ=x^p1 + (x+ 1)^p1 = [x+ (x+ 1)]A= (2x+ 1)A, A∈N Thus (2x+ 1)|aⁿ.However, 2x+ 1 is a prime number, therefore

(2x+ 1)|a Therefore

(2x+ 1)²|a² Butn >1, therefore

(2x+ 1)²|aⁿ or (2x+ 1)²|x^p1 + (x+ 1)^p1 (2) We obtain

x^p1+ (x+ 1)^p1 =x^p1+ [(2x+ 1)−x]^p1 =x^p1+ (2x+ 1)^p1+

+ µp₁

1

¶

(2x+ 1)^p1−1(−x) +...

µ p₁ p₁−1

¶

(2x+ 1) (−x)^p1−1+ (−x)^p1 =

=x^p1 + (2x+ 1)²B+ µ p₁

p₁−1

¶

(2x+ 1)x^p1−1−x^p1 =

= µ p₁

p₁−1

¶

(2x+ 1)x^p1−1+ (2x+ 1)²B, B∈Z From the relation (2), it follows that

(2x+ 1)²| µ p₁

p₁−1

¶

(2x+ 1)x^p1−1 or (2x+ 1)|

µ p₁ p₁−1

¶

x^p1−1 or (2x+ 1)|p₁x^p1−1

Since the number 2x+ 1 is prime,it is implied that (2x+ 1)|p₁ or (2x+ 1)|x.

The second case is not allowed since

2x+ 1> x k as (2x+ 1, x) = 1

Finally (2x+ 1)|p₁,i.e. p₁ = 2x+ 1, since p₁ is a prime number.

Hence p₁ =p₂.However, we have assumed that the equation does not accept integer solutions forp₁ =p₂,therefore the equation (1) and thus the equation

µp₂−1 2

¶_p1 +

µp₂−1 2

¶_p1

=aⁿ does not accept integer solutions.

W6. It is evident that

p(n) =p⁰₁(n) +p⁰₂(n) +...+p⁰_n(n) = 1 +p⁰₂(n) +...+p⁰_n−1(n) + 1 (1) However for¥_n

2

¦≤m≤nit holds.

P_m⁰ (n) =p(n−m), (2)

as it easily follows from the Ferrer‘s diagrams.

From (1) and (2) we get

p(n) = 2 + h

p⁰₂(n) +...+p⁰bⁿ_ec⁻¹(n) i

+ h

p⁰bⁿ₂c(n) +...+p⁰_n−1(n) i

=

= 2 + h

p⁰₂(n) +...+p⁰bⁿ_ec−1(n) i

+ h

p

³ n−

jn 2

k´

+...+p(1) i

(3) Ifn is an even integer, then

n− jn

2 k

= 2k−

¹2k 2

º

=k= jn

2 k

for somek∈N.

In the casen is an odd integer, it follows n−

jn 2 k

= (2k+ 1)−

¹2k+ 1 2

º

= (2k+ 1)−k=k+ 1 = jn

2 k

+ 1,

which means that

n− jn

2 k

= jn

2 k

+χ₁(n),

whereχ₁(n) denotes the principal character Dirichlet modulo 2.

This is valid since

χ₁(n) =

½ 1, if (n,2) = 1 0, if (n,2) = 0 Therefore from relation (3) we obtain the equality

p(n) = 2 + h

p⁰₂(n) +...+p⁰bⁿ₂c−1(n) i

+ h

p

³jn 2 k

+χ₁(n)

´

+...+p(1) i

,

which coincides with the equality that we wished to prove.

W7. If x= ^ab_t, dx=−^ab_t2dt, and after then we substitutet→x we give

I = Zb

a

¡x²−ab¢

ln^x_aln^x_bdx (x²+a²) (x²+b²) =

Zb

a

³¡_ab

x

¢₂

−ab

´

ln_x^b ln^a_x

³¡_ab

x

¢₂

+a²´ ³¡_ab

x

¢₂ +b²

´abdx x² = Zb

a

¡ab−x²¢

ln^x_aln^x_bdx

(x²+a²) (x²+b²) =−I, therefore I = 0 butx²−¡_a+b

2

¢₂

< x²−ab which prove that Zb

a

³

x²−¡_a+b

x

¢₂´

ln^x_aln^x_bdx (x²+a²) (x²+b²) >

Zb

a

¡x²−ab¢

ln^x_aln^x_bdx (x²+a²) (x²+b²) = 0

W8. The identity can be written in following form:

a_n= [ⁿ₂] X

k=0

(−1)^k µpn

k

¶µ(q−p)n n−2k

¶

=

¡_(p+q)n

n

¢

¡_(p+q)n

pn

¢ Xn

k=0

(−1)^k µn

k

¶µ(p+q−1)n pn−k

¶

=b_n From this we give

b_n= Xn

k=0

(−1)^k µpn

k

¶µ qn n−k

¶

Denoteα the coefficient ofxⁿ in the expressionf(x) = (1−x)^pn(1 +x)^qn, and

f(x) = µµpn

0

¶

−x µpn

1

¶ +

µpn 2

¶

x²−...

¶ µµqn 0

¶ +x

µqn 1

¶ +x²

µqn 2

¶ +...

¶

therefore α=

µpn 0

¶µqn n

¶

− µpn

1

¶µ qn qn−1

¶ +

µpn 2

¶µ qn n−2

¶

−...=

= Xn

k=0

(−1)^k µpn

k

¶µ qn n−k

¶

=b_n In another way we obtain f(x) =¡

1−x²¢_pn

(1−x)^(q−p)n,and α=

µpn 0

¶µ(q−p)n n

¶

− µpn

1

¶µ(q−p)n n−2

¶

+...=

= [ⁿ₂] X

k=0

(−1)^k µpn

k

¶µ(q−p)n n−2k

¶

=a_n.

W9. If we consider the module series:

X

n≥0

|(1−x) (1−2x)...(1−nx)|

n! ,

and if we apply it Cauchy-d1Alembert criterion, we find:

n→∞lim

¯¯

¯¯a_n+1 a_n

¯¯

¯¯= lim

n→∞

|1−(n+ 1)x|

n+ 1 =|x|

IfP|x|<1 the initial series is convergent. Also, forx= 1, this series becomes

n≥0

0,so it is convergent to.

Ifx=−1,we obtain the series P

n≥0

(n+ 1),which is divergent.

We conclude that convergence set for the initial series is C= (−1,1].

In the following, we apply the binomial series for (1 +x)^1/x,when α= _x¹, x6= 0 and x∈(−1,1).We have:

(1 +x)^1/x= 1 +X

n≥1 1x

¡₁

x −1¢ ...¡₁

x −(n−1)¢

n! n^x=

= 1 +X

n≥1

(1−x) (1−2x)...(1−(n−1)x)

n! =X

n≥0

(1−x) (1−2x)...(1−nx) n!

Now, we conclude, X

n≥0

(1−x) (1−2x)...(1−nx)

n! =





(1 +x)^1/x, x∈(−1,1)\ {0}

e, x= 0

0, x= 1

Surely lim

(n,x)→(∞,0)=e.

W10. i). Eliminatingkfixed points from 1,2, ..., n remainn−k points. So, we have (n−k)^n−k functions g:{1,2, ..., n−k} → {1,2, ..., n−k}. Because kpoints from n, can be choosen in ¡_n

k

¢posibilities, it occurs:

|F|= (n−k)^n−k µn

k

¶

ii). We shall use Stirling formula of the factorial n! =

³n e

´_n√

2πne^θ/12n, θ ∈(0,1) Ifn= 2k,we have:

µ2k k

¶

= (2k)!

(k!)² =

¡_2k

e

¢_2k√

4πke^θ1/24k hk

e

√2πke^θ2/12k i₂ ,

whereθ₁, θ₂ ∈(0,1). Therefore

µ2k k

¶

= 2^2ke^6k1

³θ1

4 −θ₂

´

√πk <2^2ke^1/24k, because ^θ₄¹ −θ₂ < ¹₄ and ^√1

πk <1.

More, becausee^1/24k < e, it occurs:

µ2k k

¶

< e·2^2k Finally,

|F|=k^k µ2k

k

¶

< e(4k)^k iii). Forn= 2k, the equation|F|= 540 becomes:

k^k µ2k

k

¶

= 540.

We notice that k= 3 is a unique solution, because by induction, k^k¡_2k

k

¢>540,(∀)k≥4.So,n= 6.

W11. The idea is to order the given polynomial with respect the powers of m. It obtains the quadratic equation with the discriminant4=x⁴(x−2)² :

x⁴m²+ 3x³m+¡

2x²+x−1¢

= 0 or

¡mx²+x+ 1¢ ¡

mx²+ 2x−1¢

= 0 Thus

x₁₂= −1±√ m+ 1

m , x₃₄= −1±√ 1−4m m Case 1.

1−m

2m = −1 +√ m+ 1

m ⇒m₁ = 5−2√ 5 Case 2.

1−m

2m = −1−√ m+ 1

m ⇒m₂ = 5 + 2√ 5 Case 3.

1−m

2m = −1 +√ 1−4m

2m ⇒m₃₄=±i√

3 are not real Case 4.

1−m

2m = −1−√ 1−4m

2m , no real solutions.

W12. The function g(x) = lnf(x) satisfies:

a). g(ax)≤ag(x)

b). g(x+y)≤g(x) +g(y)

Then, with the aid of a), we deduce that g(x) =g

µ x

x+y(x+y)

¶

≤ x

x+yg(x+y) and

g(y) =g µ y

x+y(x+y)

¶

≤ y

x+yg(x+y)

By addition, g(x) +g(y)≤g(x+y) and comparing with b) , we have g(x) +g(y) =g(x+y)

By a standard procedure,g(x) =λx, whereλ∈(0,∞)∩Q.

Finally, f(x) =e^λx.

Remark that if continuiuty is a part of the contest programme, then (0,∞)∩Qcan be replaced by (0,∞).

W13. Ifa₁ =a₂ =...=a_n then the equality holds. Suppose that between a_k, k= 1, n,are two different numbers. The functionf :R→R defined by

f(x) = Ã

1 n

Xn

k=1

a^x_k

!¹

x

is strictly increasing in this case. (see: P´olya-Szeg˝o) This implies that

à 1 n

Xn

k=1

a⁵_k

!¹

5

>

à 1 n

Xn

k=1

a⁴_k

!¹

4

which is equivalent to Ã

1 n

Xn

k=1

a⁵_k

!₄

>

à 1 n

Xn

k=1

a⁴_k

!₅

(1)

On the other hand the inequality holds:

Xn

k=1

a⁴_k> 2 n−1

X

1≤q<l≤n

a²_qa²_l (2)

because it is equivalent to

X

1≤q<l≤n

¡a²_q−a²_l¢

>0.

Now inequalities (1) and (2) imply the desired inequality (**). We have proved the inequality and we shown that equality holds if and only if a₁=a₂ =...=a_n.

W14. We use the substitution x=t^a+1a+2 and we get the equalities:

Z1

0

x^a+1

f(x)dx = a+ 1 a+ 2

Z1

0

t^a f

³ t^a+1a+2

´dt= a+ 1 a+ 2

Z1

0

x^a f

³ x^a+1a+2

´dx (3)

Since f is an increasing function and x^a+1a+2 ≥0,(∀)x∈[0,1] the inequality follows: f

³ x^a+1a+2

´

≥f(x),(∀)x∈[0,1]. Thus the following inequality holds;

a+ 1 a+ 2

Z1

0

x^a f

³ x^a+1a+2

´dx≤ a+ 1 a+ 2

Z1

0

x^a

f(x)dx (4)

Now (3) and (4) imply the result.

W15. Since

b+c= 2R(sinB+ sinC) = 4RsinB+C

2 cosB−C

2 =

= 4RcosA

2 cosB−C

2 ≤4RcosA 2, it follows that

b+c≤4RcosA 2

In an analogous way, we havec+a≤4Rcos^B₂;a+b≤4Rcos^C₂.Therefore

xⁿcosA

2 +yⁿcosB

2 +zⁿcosC 2 ≥ 1

4R[xⁿ(b+c) +yⁿ(c+a) +zⁿ(a+b)] =

= 1

4R[b(xⁿ+zⁿ) +c(yⁿ+xⁿ) +a(zⁿ+yⁿ)]≥

≥ 1 4R

³

2b(xz)ⁿ² + 2c(xy)ⁿ² + 2a(yz)ⁿ²

´

= (xy)ⁿ² sinC+(yz)ⁿ² sinA+(zx)ⁿ² sinB.

Consequently, we obtain

xⁿcosA

2 +yⁿcosB

2 +zⁿcosC

2 ≥(xy)ⁿ² sinC+ (yz)ⁿ² sinA+ (zx)ⁿ² sinB.

Remark. In same way we obtain the following stronger inequality

xⁿcosA

2+yⁿcosB

2+zⁿcosC 2 ≥

µy+z 2

¶_n

sinA+

µz+x 2

¶_n

sinB+

µx+y 2

¶_n sinC

W16. We check the inequality forn= 1, so _τ(1)¹ = 1>√

2−1,which is true. We assume that the inequality holds forn. Hence we prove that the inequality is true and for n+ 1, thus:

n+1X

k=1

1 τ(k) =

Xn

k=1

1

τ(k)+ 1

τ(n+ 1) >√

n+ 1−1 + 1 τ(n+ 1), butτ(n)<2√

n, for anyn≥1,so _τ(n+1)¹ > ₂^√1_n+1,which means that

n+1X

k=1

1

τ(k) >√

n+ 1−1 + 1 2√

n+ 1 >√

n+ 2−1 By the mathematical induction, the inequality is true for anyn≥1.

W17. Ifa=b=c= 1 then we have equality. We considerc= max{a, b, c}

and thenc≥1. We show that

(a+ 1) (b+ 1) (c+ 1)≥min{f(b), g(b)}

and similar result that

(a+ 1) (b+ 1) (c+ 1)≥min{f(a), f(b)}

We have a= _bc¹ and

(a+ 1) (b+ 1) (c+ 1) = µ1

bc+ 1

¶

(b+c) (c+ 1) = 2 + 1

bc+bc+b+c+1 b+1

c

Ifh(c) = 2 +_bc¹ +bc+b+c+¹_b +¹_c,then h⁰(c) = (b+ 1)¡

bc²−1¢ bc² evidently

h: [1,+∞)→(0,+∞). Ifb <1, then

c 1 ^√1

b +∞

h⁰ − − − − 0 + ++ + h ^2(b+1)_b ² & (b+ 1)

³√2 b + 1

´

% therefore h(c)≥g(b).Ifb≥1, then

c 1 +∞

h⁰ + + + ++ ++

h ^2(b+1)_b ² %

therefore h(c)≥f(b). W18. We have

a²ⁿ⁺³+b²ⁿ⁺³ ≥a²ⁿ⁺²b+ab²ⁿ⁺² ⇔(a−b)¡

a²ⁿ⁺¹−b²ⁿ⁺¹¢

≥0 By induction we get

a²ⁿ⁺¹+b²ⁿ⁺¹ ≥(a+b) (ab)ⁿ therefore

ab

a²ⁿ⁺³+b²ⁿ⁺³+ab ≤ ab

a²ⁿ⁺²b+ab²ⁿ⁺²+ab = 1

a²ⁿ⁺¹+b²ⁿ⁺¹+ 1 ≤

≤ 1

(a+b) (ab)ⁿ+ 1 = 1

(a+b) (ab)ⁿ+ (abc)ⁿ = 1

(ab)ⁿ(a+b+cⁿ) =

= cⁿ

a+b+cⁿ or

abc

a²ⁿ⁺³+b²ⁿ⁺³+ab ≤ cⁿ⁺¹

a+b+cⁿ ⇔ a+b+cⁿ

a²ⁿ⁺³+b²ⁿ⁺³+ab ≤cⁿ⁺¹⇔

⇔X a+b+cⁿ

a²ⁿ⁺³+b²ⁿ⁺³+ab ≤X aⁿ⁺¹

W19. We have

x²₁

¡1 +x²₁¢₂ ≤ x²₁

1 +x²₁ = 1− 1 1 +x²₁ and for k≥2

µ x_k 1 +x²₁+...+x²_k

¶₂

≤ x²_k

¡1 +x²₁+...+x²_k−1¢ ¡

1 +x²₁+...+x²_k¢ =

= 1

1 +x²₁+...+x²_k+1 − 1

1 +x²₁+...+x²_k so

Xn

k=1

µ x_k 1 +x²₁+...+x²_k

¶₂

≤ µ x₁

1 +x²₁

¶₂ +

+ Xn

k=2

Ã

1

1 +x²₁+...+x²_k−1 − 1

1 +x²₁+...+x²_k

!

≤1− 1

1 +x²₁ + 1 1 +x²₁−

− 1 1 + Pⁿ

k=1

x²_k

= Pn k=1

x²_k 1 + Pⁿ

k=1

x²_k

W20. 1). If z= cos 2x+isin 2x, then X

0≤j<k≤n

sin (2 (j+k)x) =Im



 X

0≤j<k≤n

z^j+k



= 1 2







 Xn

j=0

z^j





2

− Xn

j=0

z^2j



=

= 1 2Im

"µ zⁿ⁺¹ z−1

¶₂

− z²ⁿ⁺²−1 z²−1

#

2).

X

0≤j<k≤n

cos (2 (j+k)x) = 1 2Re

"µ

zⁿ⁺¹−1 z−1

¶₂

−z²ⁿ⁺²−1 z²−1

#

but

zⁿ⁺¹−1

z−1 = sin (n+ 1)x

sinx (cosnx+isinnx) so

X

0≤j<k≤n

sin (2 (j+k)x) = sinnxsin (n+ 1)xsin 2nx sinxsin 2x

and

X

0≤j<k≤n

cos (2 (j+k)x) = sinnxsin (n+ 1)xcos 2nx sinxsin 2x

and



 X

0≤j<k≤n

sin (2 (j+k)x)





2

+



 X

0≤j<k≤n

cos (2 (j+k)x)





2

=

= sin²nxsin²(n+ 1)x sin²xsin²2x W21. We prove by induction the inequality

Xn

k=1

x_k 1 +x_k ≥

Pn k=1

x_k 1 + Pⁿ

k=1

x_k

It is true for n= 1, for n= 2 after computation we have (x₁+x₂)x₁x₂+ 2x₁x₂ ≥0 true. We suppose true forn and we prove forn+ 1.

n+1X

k=1

x_k 1 +x_k =

Xn

k=1

x_k

1 +x_k + x_n+1 1 +x_n+1 ≥

Pn k=1

x_k 1 + Pⁿ

k=1

+ x_n+1 1 +x_n+1 ≥

n+1P

k=1

x_k 1 +ⁿ⁺¹P

k=1

x_k

the last step is the case forn= 2.

Ifx_k= _k¹s,then Pⁿ

k=1

k^s1+1 ≥

Pn k=1

1 ks

1+Pⁿ

k=1 1 ks

.In this we taken→ ∞,and we obtain the desired inequality.

W22. Fork∈ {0,1} we have equality. Using the weighted AM-GM inequality we get

µa₁ a₂

¶_k

+ k−1 k+n−1

õa₂ a₃

¶_k

+...+ µa_n

a₁

¶_k!

≥ nk

k+n−1·a₁ a₂ therefore

nk k+n−1

X µa₁ a₂

¶_k

= X

cyclic

õa₁ a₂

¶_k

+ k−1 k+n−1

õa₂ a₃

¶_k

+...+ µa_n

a₁

¶_k!!

≥ nk k+n−1

Xa₁ a₂ therefore

X µa₁ a₂

¶_k

≥Xa₁ a₂

W23. Ifx, y∈R then

¡x²+xy+y²¢_m

≤ 3^m 2

¡x^2m+y^2m¢

(1)

Form= 1 is true. We suppose true for m, and we prove form+ 1.

¡x²−xy+y²¢_m+1

≤ 3^m 2

¡x^2m+y^2m¢ ¡

x²−xy+y²¢

≤

≤ 3^m+1 2

¡x^2m+2+y^2m+2¢ We have

¡x^2m+y^2m¢ ¡

x²−xy+y²¢

≤3¡

x^2m+2+y^2m+2¢

⇔

⇔¡

x^2m+1+y^2m+1¢

(x+y) +¡

x^2m−y^2m¢ ¡

x²−y²¢

≥0 1). P

cyclic

¡x²₁−x₁x₂+x²₂¢_m

≤ P

cyclic 3^m

2

¡x^2m₁ +x^2m₂ ¢

= 3^m Pⁿ

k=1

x^2m_k 2). Q

cyclic

¡x²₁−x₁x₂+x²₂¢_m

≤¡₃m

2

¢_n Q

cyclic

¡x^2m₁ +x^2m₂ ¢

≤

¡₃m

2

¢_nÃ

n1

P

cyclic

¡x^2m₁ +x^2m₂ ¢!_n

=¡₃m

2

¢_nµ

n2

Pn k=1

x^2m_k

¶_n

=¡₃m

n

¢_nµ Pn k=1

x^2m_k

¶_n

W24. IfA(a), B(b), C(c), P(z), K¡_a+b

2

¢, L¡_b+c

2

¢, M¡_c+a

2

¢and X ¯¯¯

¯z₁−z₂ z₁+z₂

¯¯

¯¯≥

¯¯

¯¯z₁−z₂

z₁+z₂ +z₂−z₃

z₂+z₃ +z₃−z₁ z₃+z₁

¯¯

¯¯=

=

¯¯

¯¯z₁−z₂

z₁+z₂ +z₂−z₃

z₂+z₃ −z₁−z₂

z₃+z₁ −z₂−z₃ z₃+z₁

¯¯

¯¯=

=

¯¯

¯¯(z₁−z₂) (z₃−z₂)

(z₁+z₂) (z₃+z₁) +(z₁−z₂) (z₂−z₃) (z₂+z₃) (z₃+z₁)

¯¯

¯¯=

¯¯

¯¯(z₁−z₂) (z₂−z₃) (z₃−z₁) (z₁+z₂) (z₂+z₃) (z₃+z₁)

¯¯

¯¯

Ifz₁ =z−a, z₂ =z−b, z₃ =z−c,then X1

2

¯¯

¯¯

¯ a−b z−^a+b₂

¯¯

¯¯

¯≥ 1 8

¯¯

¯¯

¯

Y a−b z−^a+b₂

¯¯

¯¯

¯⇒X AB P K ≥ 1

4

Y AB P K

W25. Ifx, y >0 and x²+y²= 1,then x+y+ 1

xy ≥2 +√

2 (∗)

Proof. We have E =x+y+ 1

xy = (x+y)p

x²+y²+ 1

xy ≥2√ xyp

2xy+ 1

xy = 2√

2xy+ 1 xy.

But 1 = x² +y² ≥ 2xy, xy = t ⇒ t ≤ ¹₂ and t≤ ^√1

2 ⇒ E = 2√

2t+¹_t ≥2 +√

2 ⇔2√ 2t²−

¡2 +√ 2¢

t+ 1≥0⇔¡

t−¹₂¢ ³ t−^√1

2

´

≥0

In triangleABD we take (1)

x= AD

BD, y= AB BD

(∗)⇒ AD+AB

BD + BD²

AD·AB ≥2 +√ 2 In triangleBCD we take

(2)

x= BC

BD, y= DC BD

(∗)⇒ BC+CD

BD + BD²

BC·CD ≥2 +√ 2 Adding (1) and (2) we get

1

BD(AB+BC+CD+DA) +BD²

µ 1

AB·AD+ 1 CB·CD

¶

≥2

³ 2 +√

2

´

W26. We have 1 = Pⁿ

i=1

a^k_i ≥nⁿ sQn

i=1

a^k_i if Qn

i=1

ai=t⇒1≥n√ⁿ

t^k⇒t≤n^−nk ≤n^−1+k1,becausek≤n+ 1, therefore Xn

i=1

ai+ 1 Qn i=1

a_i

= Ã _n

X

i=1

ai

! Ã _n X

i=1

a^k_i

!ⁿ⁻¹

k

+ 1

Qn i=1

a_i

≥

≥nⁿ vu utYⁿ

i=1



nⁿ vu utYⁿ

i=1

a^k_i





n−1 k

+ 1

Qn i=1

ai

=n^{n−1k +1} Yn

i=1

ai+ 1 Qn i=1

ai

=n^{n−1k +1}t+1 t ≥

≥n^1−1k+n^nk ⇔n^{n−1k +1}t²−

³

n^1−1k +n^nk

´

t+1≥0⇔

³

n^nkt−1

´ ³

n^1−k1t−1

´

≥0 true

W27. It is a known fact that for every perfect number mit holds X

d|m

1 d = 2.

Therefore

X

d|aⁿ

1 d = 2.

Assume that

aⁿ=p^k₁¹p^k₂²...p^k_µ^µ

is the unique prime factorization of the number aⁿ.It is obvious that 1

p^k₁¹ + 1

p^k₂² +...+ 1

p^k_µ^µ <2 (1)

From Cauchy‘s arithmetic-geometric mean inequality it follows

µ

q

p^k₁¹p^k₂²...p^k_µ^µ ≥ µ

1 p^k₁¹ + ¹

p^k₂² +...+ ¹

p^kµµ

⇔ 1

p^k₂² +...+ 1

p^k_µ^µ ≥ µ

µ

q

p^k₁¹p^k₂²...p^k_µ^µ Therefore

1 p^k₁¹ + 1

p^k₂² +...+ 1 p^kµ^µ

≥ µ

a^n/µ (2)

From (1) and (2) it follows that 2> µ

a^n/µ ⇔a^n/µ> µ 2 W28. Sety= 0 in (1). Then

°°

°2f

³x 2

´

−f(x)

°°

°Y ≤θkxk^p_X