Advanced book on Mathematics Olympiad

Algebra 365

The equality holds if and only ifa=b=1, i.e.,x=0.
(T. Andreescu, Z. Feng, 101Problems in Algebra, Birkhäuser, 2001)

96.Clearly, 0 is not a solution. Solving fornyields−^4 xx 4 −^3 ≥ 1, which reduces to
x^4 + 4 x+ 3 ≤0. The last inequality can be written in its equivalent form,

(x^2 − 1 )^2 + 2 (x+ 1 )^2 ≤ 0 ,

whose only real solution isx=−1.
Hencen=1 is the unique solution, corresponding tox=−1.
(T. Andreescu)

97.Ifx=0, theny=0 andz=0, yielding the triple(x,y,z)=( 0 , 0 , 0 ).Ifx =0,
theny =0 andz =0, so we can rewrite the equations of the system in the form

1 +

1

4 x^2

=

1

y

,

1 +

1

4 y^2

=

1

z

,

1 +

1

4 z^2

=

1

x

.

Summing up the three equations leads to

(

1 −

1

x

+

1

4 x^2

)

+

(

1 −

1

y

+

1

4 y^2

)

+

(

1 −

1

z

+

1

4 z^2

)

= 0.

This is equivalent to
(
1 −

1

2 x

) 2

+

(

1 −

1

2 y

) 2

+

(

1 −

1

2 z

) 2

= 0.

It follows that 21 x= 21 y= 21 z=1, yielding the triple(x,y,z)=(^12 ,^12 ,^12 ). Both triples
satisfy the equations of the system.
(Canadian Mathematical Olympiad, 1996)

98.First, note that(x−^12 )^2 ≥0 impliesx−^14 ≤x^2 , for all real numbersx. Applying
this and using the fact that thexi’s are less than 1, we find that

logxk

(

xk+ 1 −

1

4

)

≥logxk(xk^2 + 1 )=2 logxkxk+ 1.

Therefore,

∑n

k= 1

logxk

(

xk+ 1 −

1

4

)

≥ 2

∑n

k= 1

logxkxk+ 1 ≥ 2 nn

√

lnx 2

lnx 1

·

lnx 3

lnx 2

···

lnxn

lnx 1

= 2 n.