Algebra 383

while the product of the second pair is

i^2 +im+ik+i(j− 1 )+(j− 1 )m+(j− 1 )k+(j− 1 )^2.

We can see that the first of these expressions exceeds the second bymk. This proves that if
the permutation has an inversion, then the product is not minimal. The only permutation
without inversions is the identity permutation. By Sturm’s principle, it is the permutation
for which the minimum is attained. This minimum is 1· 3 ··· 5 ···( 2 n− 1 ), as claimed.

138.Order the numbersx 1 <x 2 <···<xnand call the expression from the statement

E(x 1 ,x 2 ,...,xn). Note thatE(x 1 ,x 2 ,...,xn)>x

(^2) n
n, which shows that as the variables
tend to infinity, so does the expression. This means that the minimum exists. Assume
that the minimum is attained at the point(y 1 ,y 2 ,...,yn).Ifyn−y 1 >nthen there
exist indicesiandj,i<j, such thaty 1 ,...,yi+ 1 ,...,yj− 1 ,...,ynare still distinct
integers. When substituting these numbers intoEthe denominator stays constant while
the numerator changes by 3(yj+yi)(yj−yi− 1 ), a negative number, decreasing the value
of the expression. This contradicts the minimality. We now look at the case with no gaps:
yn−y 1 =n−1. Then there existsasuch thaty 1 =a+ 1 ,y 2 =a+ 2 ,...,yn=a+n.
We have
E(y 1 ,...,yn)=
na^3 + 3 n(n 2 +^1 )a^2 +n(n+^1 )( 22 n+^1 )a+n
(^2) (n+ 1 ) 2
4
na+n(n 2 +^1 )

a^3 +^3 (n 2 +^1 )a^2 +(n+^1 )( 22 n+^1 )a+n(n+^1 )
2
4
a+n+ 21

.

Whena=0 this is justn(n 2 +^1 ). Subtracting this value from the above, we obtain

a^3 +^3 (n 2 +^1 )a^2 +

[

(n+ 1 )( 2 n+ 1 )

2 −

n(n+ 1 )

2

]

a

a+n+ 21

> 0.

We deduce thatn(n 2 +^1 )is a good candidate for the minimum.
Ifyn−y 1 =n, then there existaandksuch thaty 1 =a,...,yk=a+k− 1 ,yk+ 1 =
a+k+ 1 ,...,yn=a+n. Then

E(y 1 ,...,yn)=

a^3 +···+(a+k− 1 )^3 +(a+k+ 1 )^3 +···+(a+n)^3

a+···+(a+k− 1 )+(a+k+ 1 )+···+(a+n)

=

∑n

j= 0 (a+j)

(^3) −(a+k) 3
∑n
j= 0 (a+j)−(a+k)

=

na^3 + 3

[

n(n+ 1 )

2 −k

]

a^2 + 3

[

n(n+ 1 )( 2 n+ 1 )

6 −k

2

]

a+

[

n^2 (n+ 1 )^2

4 −k

3

]

na+n(n 2 +^1 )−k

.