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

.