Advanced book on Mathematics Olympiad

2.1 Identities and Inequalities 37

C

A

B

o

o

O a

b

c

60

60

Figure 12

Example.LetP(x)be a polynomial whose coefficients lie in the interval[ 1 , 2 ], and let

Q(x)andR(x)be two nonconstant polynomials such thatP(x)=Q(x)R(x), withQ(x)

having the dominant coefficient equal to 1. Prove that|Q( 3 )|>1.

Solution.LetP(x)=anxn+an− 1 xn−^1 +···+a 0. We claim that the zeros ofP(x)lie

in the union of the half-plane Rez≤0 and the disk|z|<2.

Indeed, suppose thatP(x)has a zerozsuch that Rez>0 and|z|≥2. From

P(z)=0, we deduce thatanzn+an− 1 zn−^1 =−an− 2 zn−^2 −an− 3 zn−^3 −···−a 0. Dividing

through byzn, which is not equal to 0, we obtain

an+

an− 1

z

=−

an− 2

z^2

−

an− 3

z^3

−···−

a 0

zn

.

Note that Rez>0 implies that Re^1 z>0. Hence

1 ≤an≤Re

(

an+

an− 1

z

)

=Re

(

−

an− 2

z^2

−

an− 3

z^3

−···−

a 0

zn

)

≤

∣∣

∣

∣−

an− 2

z^2

−

an− 3

z^3

−···−

a 0

zn

∣∣

∣

∣≤

an− 2

|z|^2

+

an− 3

|z|^3

+···+

a 0

|z|n

,

where for the last inequality we used the triangle inequality. Because theai’s are in the

interval[ 1 , 2 ], this is strictly less than

2 |z|−^2 ( 1 +|z|−^1 +|z|−^2 +···)=

2 |z|−^2

1 −|z|−^1

.

The last quantity must therefore be greater than 1. But this cannot happen if|z|≥2,
because the inequality reduces to(|^2 z|− 1 )(|^1 z|+ 1 )>0, impossible. This proves the
claim.
Returning to the problem,Q(x)=(x−z 1 )(x−z 2 )···(x−zk), wherez 1 ,z 2 ,...,zk
are some of the zeros ofP(x). Then

|Q( 3 )|=| 3 −z 1 |·| 3 −z 2 |···| 3 −zk|.