684 Number Theoryorrn≤
r 1
1+
r 2
2+
r 3
3+···+
rn
n.
Truncate the sum on the right to the(q− 1 )st term. Sincepandqare coprime, the
numbersr 1 ,r 2 ,...,rq− 1 are a permutation of 1, 2 ,...,q−1. Applying this fact and the
AM–GM inequality, we obtain
r 1
1+
r 2
2+
r 3
3+···+
rq− 1
q− 1≥(q− 1 )(
r 1
1·
r 2
2·
r 3
3···
rq− 1
q− 1) 1 /(q− 1 )
=(q− 1 )≥rn.This proves the (weaker) inequality
r 1
1+
r 2
2+
r 3
3+···+
rn
n
≥rn,and consequently the inequality from the statement of the problem.
(O.P. Lossers)
722.Letx 1 be the golden ratio, i.e., the (unique) positive root of the equationx^2 −x− 1 =- We claim that the following identity holds:
⌊
x 1
⌊
x 1 n+1
2
⌋
+
1
2
⌋
=
⌊
x 1 +1
2
⌋
+n.If this were so, then the functionf (n)=x 1 n+^12 would satisfy the functional equation.Also, sinceα=^1 +
√ 5
2 >1,fwould be strictly increasing, and so it would provide an
example of a function that satisfies the conditions from the statement.
To prove the claim, we only need to show that
⌊
(x 1 − 1 )⌊
x 1 n+1
2
⌋
+
1
2
⌋
=n.We have
⌊
(x 1 − 1 )⌊
x 1 n+1
2
⌋
+
1
2
⌋
≤
⌊
(x 1 − 1 )(
x 1 n+1
2
)
+
1
2
⌋
=
⌊
x 1 n+n−x 1 n+x 1
2⌋
=n.Also,
n=⌊
n+2 −x 1
2⌋
≤
⌊
(x 1 − 1 )(
x 1 n−1
2
)
+
1
2
⌋
≤
⌊
(x 1 − 1 )⌊
x 1 n+