Advanced book on Mathematics Olympiad

Algebra 385

Applying it fork=n=3,r 1 =r 2 =r 3 =3, and the numbersa 11 =a,a 12 =1,a 13 =1,
a 21 =1,a 22 =b,a 23 =1,a 31 =1,a 32 =1,a 33 =c, we obtain

(a+b+c)≤(a^3 + 1 + 1 )

(^13)
( 1 +b^3 + 1 )
(^13)
( 1 + 1 +c)
(^13)
.
We thus have
(a^3 + 1 + 1 )( 1 +b^3 + 1 )( 1 + 1 +c^3 )≥(a+b+c)^3 ,
and the inequality is proved.
(USA Mathematical Olympiad, 2004, proposed by T. Andreescu)
141.Letxi,i= 1 , 2 ,...,n,xi>0, be the roots of the polynomial. Using the relations
between the roots and the coefficients, we obtain
∑
x 1 x 2 ···xm=

(

n

m

)

and

∑

x 1 x 2 ···xp=

(

n

p

)

.

The generalized Maclaurin inequality

m

√∑

x 1 x 2 ···xm

(n

m

) ≥ p

√∑

x 1 x 2 ···xp

(n

p

)

thus becomes equality.∑ This is possible only ifx 1 = x 2 = ··· = xn. Since

x 1 x 2 ···xm=

(n

m

)

, it follows thatxi=1,i= 1 , 2 ,...,n, and henceP(x)=(x− 1 )n.

(Revista Matematica din Timi ̧soara ̆ (Timi ̧soara Mathematics Gazette), proposed by

T. Andreescu)

142.The idea of the solution is to reduce the inequality to a particular case of the Huygens

inequality,

∏n

i= 1

(ai+bi)pi≥

∏n

i= 1

apii+

∏n

i= 1

bpii,

which holds for positive real numbersp 1 ,p 2 ,...,pn,a 1 ,a 2 ,...,an,b 1 ,b 2 ,...,bnwith

p 1 +p 2 +···+pn=1.

To this end, start with

n−xi

1 −xi

= 1 +

n− 1

x 1 +···+xi− 1 +xi+ 1 +···+xn

and apply the AM–GM inequality to get

n−xi

1 −xi

≤ 1 +

1

n−√ (^1) x 1 ···xi− 1 xi+ 1 ···xn.