Thread: Harmonic numbers

I don't think I gave this problem before.

Let be an injection. Show that

with equality only if is the embedding of in .

I guess you mean , not ?

Spoiler:

The square suggests Cauchy-Schwarz...

.

Then note that (if we order the values of as , then by induction we have since the sequence is strictly increasing).

Plugging this inequality in the above one and simplifying yields the inequality we are looking for.

Assume there is equality. Then there is equality in Cauchy-Schwarz inequality, which implies that, for some , for , hence . Since is integer-valued, we must have . And there is also equality in the other inequality : , which entails immediately. Thus, for .

Originally Posted by Laurent

I guess you mean , not ?

Haha, yes! I was just having this discussion with friends a few days ago; they claimed that unambiguously meant the set of positive integers, while I claimed that was sometimes included... I guess you can never be too precise.

Nice solution! It's completely different from mine, I'll post it a bit later.

Alright, sorry for the delay; here's my solution.

Suppose is such that the sum on the right is minimal. Then we can assume to be a permutation of ; indeed, order the values in increasing order and replace them respectively by ; we minimize the sum in such a way. Now suppose there exist with and ; suppose we switch the values of in the sum. The difference between the two sums is , meaning we have minimized the sum some more, contradicting the hypothesis. Therefore .