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Problem 11012

(American Mathematical Monthly, Vol.110, May 2003) Proposed by C. Popescu (Belgium).

Given a positive integer n, find the minimum value of x31+ · · · + x3n

x1+ · · · + xn

subject to the condition that x1, . . . , xn be distinct positive integers.

Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.

We will prove that the minimum value of the function

F(x1, x2,· · · , xn) =x31+ · · · + x3n

x1+ · · · + xn

in the set An = {0 < x1< x2<· · · < xn : xk∈ N} is

F(1, 2, · · · , n) =13+ 23+ · · · + n3 1 + 2 + · · · + n =

n(n+1)

2

2 n(n+1)

2

=n(n + 1)

2 .

The statement is trivial for n = 1, so assume that n ≥ 2. Note that the minimum value has to be attained in the finite set An∩ {|xn| ≤ n2} because for xn > n2

x31+ · · · + x3n

x1+ · · · + xn > x3n

n· xn = x2n

n > n3> n(n + 1)

2 .

Let (x1, x2, . . . , xn) be a minimum point then it suffices to show that

xn− xn−1= xn−1− xn−2= · · · = x2− x1= x1− x0= 1 (x0= 0).

If these conditions are not satisfied then there is an integer 1 ≤ r ≤ n such that the numbers xn−r+1,· · · , xnare consecutive whereas xn−r+1− xn−r >1. This gives us a contradiction: the value can be lowered by extending the sequence of consecutive numbers, that is by replacing xn−r+1+ k with xn−r+ k + 1 (smaller) in xn−r+k+1 for k = 0, . . . , r − 1

F(x1,· · · , xn−r+1,· · · , xn−r+1+ r − 1) > F (x1,· · · , xn−r+ 1, · · · , xn−r+ r).

The above inequality holds because the function

fr(x) =

n−r

X

k=1

x3k+

r

X

k=1

(x + k)3

n−r

X

k=1

xk+

r

X

k=1

(x + k)

is strictly increasing for x ≥ xn−r: the derivative is positive if

3 ·

r

X

k=1

(x + k)2·

n−r

X

k=1

xk+

r

X

k=1

(x + k)

!

> r

n−r

X

k=1

x3k+

r

X

k=1

(x + k)3

! .

Since

3 ·

r

X

k=1

(x + k)2·

n−r

X

k=1

xk ≥ 3 · r ·

n−r

X

k=1

x2· xk ≥ r ·

n−r

X

k=1

x3k,

(2)

it remains to prove that

3 ·

r

X

k=1

(x + k)2·

r

X

k=1

(x + k) > r ·

r

X

k=1

(x + k)3.

This follows by noting that the function

gr(x) = 3 ·

r

X

k=1

(x + k)2·

r

X

k=1

(x + k) − r ·

r

X

k=1

(x + k)3

is positive for x ≥ 0. In fact

gr(0) = 3 ·

r

X

k=1

k2·

r

X

k=1

k− r

r

X

k=1

k3

= 3 ·r(r + 1)(2r + 1)

6 ·r(r + 1)

2 − r · r(r + 1) 2

2

= (r + 1) · r(r + 1) 2

2

>0 and its derivative is positive for x ≥ 0

gr(x) = 6 ·

r

X

k=1

(x + k) ·

r

X

k=1

(x + k) + 3 · r ·

r

X

k=1

(x + k)2− 3 · r ·

r

X

k=1

(x + k)2

= 6 ·

r

X

k=1

(x + k)

!2

>0.



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