On variations of a claim due to Kaneko in terms of Lehmer means In this post (now cross posted as this question on MathOverflow with identificator 362866), for a tuple of positive real numbers $\mathbb{x}=(x_1,x_2,\ldots,x_n)$ we denote its corresponding Lehmer mean as $L_q(\mathbb{x})$, where $q>0$. This is an important example of mean, the Wikipedia article dedicated to this mean is Lehmer mean that includes a section for Applications. As aside comment I don't know if any definition of mean enjoys of the feature explained in this section of applications with relation to signal processing.
Also we denote the sum of divisors function as $$\sigma(n)=\sum_{1\leq d\mid n}d$$ for integers $n\geq 1$.
The idea of the post was to combine this definition of Lehmer mean with an equivalent formulation of the Riemann hypothesis, I refer the last paragraph of [1] (Kaneko's claim for a suitable choice of the integer $n$).
From here, my belief that there should be an integer $n_0>1$ such that $\forall n\geq n_0$ the following inequality holds $$\sigma(n)<\exp\left(\frac{n}{L_q(1,\ldots,n)}\right)\log\left(\frac{n}{L_q(1,\ldots,n)}\right)\tag{1}$$
with $q>0$ ((!) this is false for $0<q<1$, see the answer). 
Fact (Is false for $0<q<1$, see the answer below). We've from the theory of Lehmer mean that we recover the inequality of Kaneko as $q$ (or $|q|$) tends to $0^{+}$.

Question. I would like to know what work can be done to get an inequality $(1)$ for a very small $|q|$ (I mean very close to $0$) that holds $\forall n\geq n_0$ for your suitable choice of $n_0>1$ (and your $q$, what I want is $|\sum_{1\leq k\leq n}\frac{1}{k}-n/L_q(1,\ldots,n)|$ small). Many thanks.

I emphasize that I'm asking what work can be done to prove an example for one of those inequalities $(1)$ for a very small quantity $q>0$. Optionally feel free to add your feedback about if this type of inequalities and combinations can be potentially interesting.
Final remarks: 1) My interpretation was for the $RHS$ of the mentioned inequality (from the article of Lagarias) due to Kaneko. I don't interpret (if it is interesting) in a similar way the $LHS$ that is the sum of divisors function (I don't know if it is feasible*). 2) If I understand well the veracity of some equivalences to the Riemann hypothesis, that were stated as inequalities involving certain arithmetic functions, depends (counterexamples) on certain integer sequences. And I don't know if it is feasible to create a mean that enjoys the same feature than the Lehmer mean (the mentioned in first paragraph, see also the link power mean from the cited Wikipedia's article for Generalized mean).
References:
[1] Jeffrey C. Lagarias, An Elementary Problem Equivalent to the Riemann Hypothesis, The American Mathematical Monthly, 109, No. 6 (2002), pp. 534-543.
[2] P. S. Bullen, Handbook of Means and Their Inequalities, Springer, (1987).
 A: This is a partial answer.
This answer proves the following two claims :
Claim 1 : It is false that if $q$ is a fixed number satisfying $0\lt q\lt 1$, then there is an integer $n_0>1$ such that $(1)$ holds $\forall n\geq n_0$.
Claim 2 : If $q$ is a fixed number satisfying $-1\lt q\lt 0$, then
$$\lim_{n\to\infty}\frac{n}{L_q(1,\ldots,n)}=\infty$$

Claim 1 : It is false that if $q$ is a fixed number satisfying $0\lt q\lt 1$, then there is an integer $n_0>1$ such that $(1)$ holds $\forall n\geq n_0$.
Proof :
For $0\lt q\lt 1$, we get 
$$\sum_{k=1}^{n}k^q\gt \int_{0}^{n}x^qdx=\frac{n^{q+1}}{q+1}$$
and
$$\sum_{k=1}^{n}k^{q-1}\lt 1+\int_{1}^{n}x^{q-1}dx=\frac{n^q-1+q}{q}$$
It follows from these that
$$\frac{n}{L_q(1,\ldots,n)}=\frac{n\sum_{k=1}^{n}k^{q-1}}{\sum_{k=1}^{n}k^q}\lt\frac{n\cdot\frac{n^q-1+q}{q}}{\frac{n^{q+1}}{q+1}}=\frac{(n^q-1+q)(q+1)}{n^{q}q}$$
Let $f(x)=\frac{(x^q-1+q)(q+1)}{x^{q}q}$. Then, we have $f'(x)= \frac{1-q^2}{ x^{q+1}}\gt 0$. So, $f(x)$ is increasing with
$$\lim_{x\to\infty}f(x)=\lim_{x\to\infty}\frac{q+1}{q}\bigg(1+\frac{q-1}{x^q}\bigg)=\frac{q+1}{q}$$
from which we have
$$\frac{n}{L_q(1,\ldots,n)}\lt f(n)\lt\frac{q+1}{q}$$
So, we get
$$\small\exp\left(\frac{n}{L_q(1,\ldots,n)}\right)\log\left(\frac{n}{L_q(1,\ldots,n)}\right)\lt \exp\left(\frac{q+1}{q}\right)\log\left(\frac{q+1}{q}\right)=(\text{a constant})$$
Considering $\sigma(2^m)=2^{m+1}-1$, we see that the conjecture is false. $\quad\square$

Claim 2 : If $q$ is a fixed number satisfying $-1\lt q\lt 0$, then
$$\lim_{n\to\infty}\frac{n}{L_q(1,\ldots,n)}=\infty$$
Proof : 
For $-1\lt q\lt 0$, we get
$$\sum_{k=1}^{n}k^q\lt 1+\int_{1}^{n}x^qdx=\frac{n^{q+1}+q}{q+1}$$
and
$$\sum_{k=1}^{n}k^{q-1}\gt \int_{1}^{n+1}x^{q-1}dx=\frac{(n+1)^q-1}{q}$$
It follows from these that
$$\frac{n}{L_q(1,\ldots,n)}=\frac{n\sum_{k=1}^{n}k^{q-1}}{\sum_{k=1}^{n}k^q}\gt\frac{n\frac{(n+1)^q-1}{q}}{\frac{n^{q+1}+q}{q+1}}=\frac{n((n+1)^q-1)(q+1)}{qn^{q+1}+q^2}$$
Finally, we get
$$\lim_{n\to\infty}\frac{n((n+1)^q-1)(q+1)}{qn^{q+1}+q^2}=\lim_{n\to\infty}\frac{q+1}{q}\cdot\frac{\frac{1}{(1+\frac 1n)^{-q}}-n^{-q}}{1+\frac q{n^{q+1}}}=\infty$$
