Algebra 403180.The hypothesis of the problem concerns the coefficientsamanda 0 , and the conclusion
is about a zero of the polynomial. It is natural to write the Viète relations for the two
coefficients,
am
an
=(− 1 )m∑
x 1 x 2 ···xm,
a 0
an
=(− 1 )nx 1 x 2 ···xn.Dividing, we obtain
∣
∣∣
∣∑ 1
x 1 x 2 ···xm∣
∣∣
∣=
∣
∣∣
∣
am
a 0∣
∣∣
∣>
(
n
m)
.
An application of the triangle inequality yields
∑ 1
|x 1 ||x 2 |···|xm|>
(
n
m)
.
Of the absolute values of the zeros, letαbe the smallest. If we substitute all absolute
values in the above inequality byα, we obtain an even bigger left-hand side. Therefore,
(
n
m)
1
αn−m>
(
n
m)
.
It follows thatα<1, and hence the corresponding zero has absolute value less than 1,
as desired.
(Revista Matematica din Timi ̧soara ̆ (Timi ̧soara Mathematics Gazette), proposed by
T. Andreescu)
181.Letf(x)=P′(x)
P(x)=
1
x−x 1+
1
x−x 2+···+
1
x−xn.
First, note that from Rolle’s theorem applied toP(x)=e−kxf(x)it follows that all roots
of the polynomialP′(x)−kP (x)are real. We need the following lemma.Lemma.If for somej,y 0 andy 1 satisfyy 0 <xj<y 1 ≤y 0 +δ(P),theny 0 andy 1 are
not zeros offandf(y 0 )<f(y 1 ).Proof.Letd=δ(P). The hypothesis implies that for alli,y 1 −y 0 ≤d≤xi+ 1 −xi.
Hence for 1≤i≤j−1 we havey 0 −xi ≥y 1 −xi+ 1 >0, and so 1/(y 0 −xi)≤
1 /(y 1 −xi+ 1 ); similarly, forj≤i≤n−1 we havey 1 −xi+ 1 ≤y 0 −xi<0 and again
1 /(y 0 −xi)≤ 1 /(y 1 −xi+ 1 ).
Finally,y 0 −xn< 0 <y 1 −x 1 ,so1/(y 0 −xn)< 0 < 1 /(y 1 −x 1 ), and the result
follows by addition of these inequalities.