Junior problems - Phần 1 pps

Junior problems

J163. Let a, b, c be nonzero real numbers such that ab + bc + ca ≥ 0. Prove that

ab

a

2 + b

2

+

bc

b

2 + c

2

+

ca

c

2 + a

2

≥ −

1

2

.

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

Solution by Ercole Suppa, Teramo, Italy

We have

X

cyc

ab

a

2 + b

2

=

X

cyc



ab

a

2 + b

2

+

1

2



−

3

2

=

X

cyc

(a + b)

2

2 (a

2 + b

2)

−

3

2

≥

X

cyc

(a + b)

2

2 (a

2 + b

2 + c

2)

−

3

2

=

2

￾

a

2 + b

2 + c

2

+ 2(ab + bc + ca)

2 (a

2 + b

2 + c

2)

−

3

2

= 1 +

ab + bc + ca

a

2 + b

2 + c

2

−

3

2

=

ab + bc + ca

a

2 + b

2 + c

2

−

1

2

≥ −

1

2

where in the last step we have used the fact that ab + bc + ca ≥ 0.

Also solved by Arkady Alt, San Jose, California, USA; Daniel Lasaosa, Universidad P´ublica

de Navarra, Spain; Perfetti Paolo, Dipartimento di Matematica, Universit`a degli studi di Tor

Vergata Roma, Italy; Prithwijit De, HBCSE, India; Andrea Ligori, Universit`a di Roma “Tor

Vergata”, Italy; Piriyathumwong P., Bangkok, Thailand.

Mathematical Reflections 4 (2010)

J164. If x and y are positive real numbers such that 

x +

√

x

2 + 1y +

p

y

2 + 1

= 2011, find the

minimum possible value of x + y.

Proposed by Neculai Stanciu, “George Emil Palade”, Buzau, Romania

First solution by Michel Bataille, France The required minimum value is √

2010

2011 .

Write x = sinh(a) and y = sinh(b) where a = ln(x+

√

x

2 + 1) > 0 and b = ln(y+

p

y

2 + 1) > 0.

From the hypothesis, we have a + b = ln(2011) and using a known formula,

x + y = sinh(a) + sinh(b) = 2 sinh 

a + b

2



cosh 

a − b

2



≥ 2 sinh 

a + b

2



= 2 sinh(ln(√

2011)

where the inequality follows from cosh(t) ≥ 1 for all t and sinh(u) > 0 for u > 0.

Since 2 sinh(ln(√

2011) = √

2011 − √

1

2011 = √

2010

2011 , we obtain

x + y ≥

2010

√

2011

.

Clearly equality holds when a = b (since cosh(0) = 1), that is, when x = y. The result follows.

Second solution by the authors

Let z = x +

√

x

2 + 1. We have z > 0 and (1) x =

z

2−1

2z

. From hypothesis y +

p

y

2 + 1 = 2011

z

,

we get (2) y =

20112−z

2

2·2011·z

. From (1) and (2),

x + y =

z

2 − 1

2z

+

20112 − z

2

2 · 2011 · z

=

2010

2 · 2011 

z +

2011

z



≥

2010

2011r

z ·

2011

z

.

The equality occurs for z

2011

z

or equivalently z

2 = 2011. Then from (1) and

(2)

we obtain

x = y =

2010

2

√

2011

=

1005

√

2011

.

So min(x + y) = √

2010

2011 .

Also solved by Arkady Alt, San Jose, California, USA; Francisco Javier Garcia Capitan, Spain;

Ercole Suppa, Teramo, Italy; Daniel Lasaosa, Universidad P´ublica de Navarra, Spain; Perfetti

Paolo, Dipartimento di Matematica, Universit`a degli studi di Tor Vergata Roma, Italy.

Mathematical Reflections 4 (2010) 2

J165. Find all triples (x, y, z) of integers satisfying the system of equations

(

x

2 + 1y

2 + 1

+

z

2

10 = 2010

(x + y)(xy − 1) + 14z = 1985.

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

Solution by Arkady Alt, San Jose, California, USA

Note that z = 10k for some integer k because z

2

10

= 2010 −

￾

x

2 + 1
￾y

2 + 1

is an integer. Let

p = x + y and q = xy − 1. Then

￾

x

2 + 1
￾y

2 + 1

= x

2

y

2 + x

2 + y

2 + 1 = (xy − 1)2 + (x + y)

2 = p

2 + q

2

and the system becomes



p

2 + q

2 + 10k

2 = 2010

pq + 140k = 1985 ⇐⇒ 

p

2 + q

2 = 2010 − 10k

2

pq = 1985 − 140k

(1)

Since (p − q)

2 = 2010 − 10k

2 − 2 (1985 − 140k) = −10 (k − 14)2

then only k = 14 can provide

solvability to (1). And for k = 14, (1) becomes 

p

2 + q

2 = 50

pq = 25 ⇐⇒ p = q = 5.

Hence, 

x + y = 5

xy = 4 ⇐⇒ 

x = 4

y = 1 or 

x = 1

y = 4 and triples (5, 1, 140),(1, 5, 140) are all

integer solutions of the original system in integers.

Also solved by Daniel Lasaosa, Universidad P´ublica de Navarra, Spain; Piriyathumwong P.,

Bangkok, Thailand.

Mathematical Reflections 4 (201