Advanced book on Mathematics Olympiad

386 Algebra

Multiplying allninequalities gives

∏n

i= 1

(

n−xi

1 −xi

)

≤

∏n

i= 1

(

1 +

1

n−√ (^1) x 1 ···xi− 1 xi+ 1 ···xn

)

.

Thus we are left to prove

∏n

i= 1

(

1 +

1

xi

)

≥

∏n

i= 1

(

1 +

1

n−√ (^1) x 1 ···xi− 1 xi+ 1 ···xn

)

.

This inequality is a product of the individual inequalities

∏

j   =i

(

1 +

1

xj

)

≥

⎛

⎝ 1 +n− 1

√√

√

√

∏

j   =i

1

xi

⎞

⎠

n− 1

,j= 1 , 2 ,...,n.

Each of these is Huygens’ inequality applied to the numbers 1, 1 ,...,1 andx^11 ,...,xi^1 − 1 ,
1
xi+ 1 ,...,xn, withp^1 =p^2 = ··· =pn=

1

n− 1.

(Crux Mathematicorum, proposed by W. Janous)

143.We will use the following inequality of Aczèl: Ifx 1 ,x 2 ,...,xm,y 1 ,y 2 ,...,ymare
real numbers such thatx 12 >x 22 +···+xm^2 , then

(x 1 y 1 −x 2 y 2 −···−xmym)^2 ≥(x^21 −x 22 −···−xm^2 )(y 12 −y 22 −···−y^2 m).

This is proved in the following way. Consider

f(t)=(x 1 t+y 1 )^2 −

∑m

i= 2

(xit+yi)^2

and note thatf(−yx^11 )≤0. It follows that the discriminant of the quadratic functionf(t)
is nonnegative. This condition that the discriminant is nonnegative is basically Aczèl’s
inequality.
Let us return to the problem. It is clear thata^21 +a 2 +···+a^2 n−1 andb^21 +b 22 +···+b^2 n− 1
have the same sign. If

1 >a 12 +a^22 +···+an^2 or 1>b^21 +b^22 +···+bn^2 ,

then by Aczèl’s inequality,

( 1 −a 1 b 1 −···−anbn)^2 ≥( 1 −a^21 −a 22 −···−an^2 )( 1 −b 12 −b^22 −···−b^2 n),

which contradicts the hypothesis. The conclusion now follows.
(USA Team Selection Test for the International Mathematical Olympiad, proposed
by T. Andreescu and D. Andrica)

144.The solution is based on the Muirhead inequality.