We will prove that the minimum value of the function F(x1, x2

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 : x^k∈ 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− x_n−1= x_n−1− x_n−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 x_n−r+1,· · · , xⁿare consecutive whereas x_n−r+1− x_n−r >1. This gives us a contradiction: the value can be lowered by extending the sequence of consecutive numbers, that is by replacing x_n−r+1+ k with x_n−r+ k + 1 (smaller) in x_n−r+k+1 for k = 0, . . . , r − 1

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

The above inequality holds because the function

f^r(x) =

n−r

X

k=1

x³k+

r

X

k=1

(x + k)³

n−r

X

k=1

x^k+

r

X

k=1

(x + k)

is strictly increasing for x ≥ x_n−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

x^k ≥ 3 · r ·

n−r

X

k=1

x²· x^k ≥ r ·

n−r

X

k=1

x³k,

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

g^′r(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.