Junior problems - Phần 2 potx

Junior problems

J175. Let a, b ∈ (0,

π

2

) such that sin2 a + cos 2b ≥

1

2

sec a and sin2

b + cos 2a ≥

1

2

sec b. Prove that

cos6

a + cos6

b ≥

1

2

.

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

Solution by Prithwijit De, HBCSE, India

We will use the following well-known trigonometric identities

(a) sin2 x = 1 − cos2 x,

(b) cos 2x = 2 cos2 x − 1,

(c) sec x =

1

cos x

.

The inequalities can be written as

2 cos2

b cos a − cos3

a ≥

1

2

(1)

and

2 cos2

a cos b − cos3

b ≥

1

2

. (2)

The signs of the inequalities are preserved because cos x is positive when x ∈



0,

π

2



. Now by

squaring both sides of (1) and (2) and adding them we get

cos6 a + cos6

b ≥

1

2

.

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; Tigran Hakobyan, Armenia.

Mathematical Reflections 6 (2010) 1

J176. Solve in positive real numbers the system of equations

(

x1 + x2 + · · · + xn = 1

1

x1

+

1

x2

+ · · · +

1

xn

+

1

x1x2···xn

= n

3 + 1.

Proposed by Neculai Stanciu, George Emil Palade Secondary School, Buzau, Romania

Solution by Tigran Hakobyan, Armenia

We have

1

x1

+

1

x2

+ · · · +

1

xn

≥

n

2

x1 + x2 + · · · + xn

= n

2

and

1

x1x2 · · · xn

≥

1

￾

x1+x2+···+xn

n

n = n

n

.

Thus,

n

3 + 1 ≥ n

n + n

2

which implies that n ≤ 2. If n = 1 we get a contradiction. For n = 2 we get

(

x1 + x + 2 = 1

1

x1

+

1

x2

+

1

x1x2

= 9

which is (n, x1, x2) ∈

￾2,

1

3

,

2

3

,

￾

2,

2

3

,

1

3

.

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; Lorenzo Pascali, Universit`a di Roma “La Sapienza”, Roma, Italy.

Mathematical Reflections 6 (2010

J177. Let x, y, z be nonnegative real numbers such that ax + by + cz ≤ 3abc for some positive real

numbers a, b, c. Prove that

r

x + y

2

+

r

y + z

2

+

r

z + x

2

+

√4 xyz ≤

1

4

(abc + 5a + 5b + 5c).

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

Solution by the author

From the given condition,

3a ≥

ax

bc +

y

c

+

z

b

3b ≥

x

c

+

by

ca

+

z

a

3c ≥

x

b

+

y

a

+

cz

ab .

Then

3(a + b + c) ≥

x + y

c

+

y + z

a

+

z + x

b

+



ax

bc +

by

ca

+

cz

ab

hence

abc + 5(a + b + c) ≥



x + y

c

+ 2c



+



y + z

a

+ 2a



+



z + x

b

+ 2b



+



abc +

ax

bc +

by

ca

+

cz

ab

≥ 2

p

2(x + y) + 2p

(y + z) + 2p

(z + x) + 4√4 xyz

and the conclusion follows. The equality holds if and only if x+y = 2c

2

, y+z = 2a

2

, z+x = 2b

2

and ax

bc =

by

ca =

cz

ab = abc. This implies b

2

c

2 = c

2a

2 = 2c

2

, c2a

2 + a

2

b

2 = 2a

2

, a2

b

2 + b

2

c

2 = 2b

2

,

that is b

2 + a

2 = c

2 + b

2 = a

2 + c

2 = 2, implying a = b = c = 1 and x = y = z = 1. If

a = b = c = x = y = z = 1, the equality, as well as the condition of the problem, hold

Mathematical Reflections 6 (2010) 3