Advanced book on Mathematics Olympiad

378 Algebra

x

1 −^1 n

1 +x

1 −^1 n

2 +x

1 −^1 n

j− 1 +x

1 −^1 n

j+ 1 +···+x

1 −^1 n

n

n− 1

≥

(

(x 1 x 2 ···xj− 1 xj+ 1 ···xn)

n−n 1 )n−^11

=(x 1 x 2 ···xj− 1 xj+ 1 ···xn)

n^1

=x

−^1 n

j.

This completes the proof.
(Romanian Team Selection Test for the International Mathematical Olympiad, 1999,
proposed by V. Cârtoaje and Gh. Eckstein)

130.First solution: Note that the triple(a,b,c)ranges in the closed and bounded set
D={(x,y,z)∈R^3 | 0 ≤x, y, z≤ 1 ,x+y+z= 1 }. The functionf (x, y, z)=
4 (xy+yz+xz)− 9 xyz−1 is continuous; hence it has a maximum onD. Let(a,b,c)
be a point inDat whichfattains this maximum. By symmetry we may assume that
a≥b≥c. This immediately impliesc≤^13.
Let us apply Sturm’s method. Suppose thatb<a, and let 0<x<a−b. We show
thatf(a−x, b+x,c) > f(a,b,c). The inequality is equivalent to

4 (a−x)(b+x)− 9 (a−x)(b+x)c > 4 ab− 9 abc,

or

( 4 − 9 c)((a−b)x−x^2 )> 0 ,

and this is obviously true. But this contradicts the fact that(a,b,c)was a maximum.
Hencea=b. Thenc= 1 − 2 a, and it suffices to show thatf (a, a, 1 − 2 a)≤0.
Specifically, this means

4 a^2 − 8 a( 1 − 2 a)− 9 a^2 ( 1 − 2 a)− 1 ≤ 0.

The left-hand side factors as−( 1 − 2 a)( 3 a− 1 )^2 =−c( 3 a− 1 )^2 , which is negative or
zero. The inequality is now proved. Moreover, we have showed that the only situations
in which equality is attained occur when two of the numbers are equal to^12 and the third
is 0, or when all three numbers are equal to^13.

Second solution: A solution is possible using the Viète relations. Here it is. Consider the
polynomial

P(x)=(x−a)(x−b)(x−c)=x^3 −x^2 +(ab+bc+ca)x−abc,

the monic polynomial of degree 3 whose roots area, b, c. Becausea+b+c=1, at
most one of the numbersa, b, ccan be equal to or exceed^12. If any of these numbers is
greater than^12 , then

P

(

1

2

)

=

(

1

2

−a

)(

1

2

−b

)(

1

2

−c

)

< 0.