Proof of Theorem 2.1.

(1)
We first prove that the point where the maximum of is attained lies on the boundary of the hypercube and moreover, it is a vertex. This result is the subject of Lemma 3.1. We then find the configuration where the maximum is realized.
Lemma 3.1.
The function attains its maximum at the point if and only if for all
Proof of Lemma 3.1.
Since is continuous on the compact interval , there is a point where attains its maximum. If is an interior point of , then for all therefore
which implies
for all However, if , then which clearly is not the maximum of Consequently, lies on the boundary of . Due to symmetry and since there exist and such that
If then For this case, consider the function defined by
If the point where the maximum of is attained is interior to , in virtue of Fermat's theorem, we deduce that
for all This is equivalent to
hence
A simple computation shows that
and for this configuration we have
Let us define the function as
and prove it is increasing. Indeed, one finds
where , and Since it follows that so is increasing and the upper bound is
This finally proves that of the numbers are equal to while the other are equal to as anticipated. This ends the proof of Lemma 3.1.
The only thing to be done is to find the value of for which the expression
attains its maximum.
To do this, consider the function defined by
and find the points where the maximum of is attained in the interval .
The critical points of are found from the equation
so they satisfy
As seen in the definition of the Stolarsky mean for this case,
It is finally found that has a single critical point
which (fortunately) is contained in the interior of
Taking into account that the second derivative of is
the extremal point is a point of maximum for , and also the function is decreasing on the interval . Because , we obtain for , and for Finally, this means that is increasing on and decreasing on .
We conclude that
The maximum of (3.13) is then attained when takes one of the values and , where
The value of this is to be called from now on.
Remark 3.2.
Because in our case
the Stolarsky mean satisfies the strict inequality , so

(2)
Using the properties of the integer part , we obtain
so
It is then enough to work out the limit
On the other hand we have
Due to symmetry the partial derivatives are equal, so the desired limit is
Taking the limit in (3.23), we obtain that the limit of as is confined to the interval
Proof of Theorem 2.3.
Considering and in Theorem 2.1, we obtain
Out of here, we can immediately obtain the best constant for which
Following the steps mentioned before, the function gets the maximum only when
where , or .
This proves that the following inequality holds:
so the best constant will be
Remark 3.3.
Although appealing, a result involving arbitrary powers would depend on which the exact value of is (out of the two possibilities). At the same time, the power on the righthandside can only be obtained for
Proof of Theorem 2.4.
To ease the notations we write and The following relation holds:
Using the notation the limit can be written as
Since the denominator converges to it only remains to examine the limit
which can be written as
It can be proven that the two terms of (3.36) converge to finite limits, and analyze each. From the hypothesis so the limit of the first term is
while second term can be written as
Since
the same argument as above can be used to obtain
where
In the end we obtain