By specialization of an inequality I can write $$2 \sum_{k=1}^{n-1} \frac{1}{d_{k}} \sum_{l=k+1}^{n} \frac{1}{d_{l}}\leq 2\frac{\sigma_0(n)-1}{\sigma_0(n)}\cdot \left( \frac{\sigma(n)}{n} \right)^2, $$ where $1=d_1<d_2\ldots<d_{\sigma_{0}(n)}=n$ are the divisors of a positive integer $n>1$, $\sigma_0(n)$ is the number of such divisors and $\sigma(n)$ is the sum of divisor function. On the other hand $$\sum_{k=1}^{n-1} \frac{1}{d_{k}} \sum_{l=k+1}^{n} \frac{1}{d_{l}}=2 \left( \frac{\sigma(n)}{n} \right)^2-\frac{\sigma_2(n)}{n},$$ where $\sigma_2(n)=\sum_{d\mid n}d^2$. Thus by comparison of such formulas we can deduce $$(\sigma(n))^2\leq \sigma_0(n)\sigma_2(n).$$ After I've tried a comparison with my computer, known the average orders of these functions and involving also a different arithmetics function to adjust a conjecture. A graph for $$\frac{\sigma_0(n)\sigma_2(n)}{(\sigma(n))^2H_n}, $$ where $H_n=1+1/2+\cdot+1/n$ is the nth harmonic number, is below

Question. Can you say if is there limit as $n\to\infty$? Can do you explain, a heuristic, why the erratic behaviour of the graph at first? Thanks in advance.
Your inequality $$(\sigma(n))^2\leq \sigma_0(n)\sigma_2(n)$$ is the Cauchy-Schwarz inequality in $\mathbb R^{\sigma_0(n)}.$ One of the vectors used in Cauchy Schwarz has all entries equal to $1,$ the other has all the divisors of $n$ as the entries.
Spanish Primorial
I tried it for the primorials, it would appear that there is no upper bound, even when dividing by your $H_n.$
Here is the same computer run, but i have saved space by not printing m or its factorization, just the largest prime dividing m.
edit: I understand now: your function ( without the $H_n$) is multiplicative in the number theory sense. This means that multiplying by a new prime $p$ multiplies the "ratio" above by the function applied to $p$ itself, which is approximately $2.$ So the number I call "ratio" above is roughly doubling each time. Let $$ f(n) =\frac{\sigma_0(n)\sigma_2(n)}{(\sigma(n))^2}. $$
Alright: from Merten's Theorem, if prime number $k$ is called $p,$ so that $p = p_k,$ the "cumulative product" is approximately a constant times $$ \frac{2^k}{\log^2 p} $$