An extended comment that has a chance of being useful.

First, note that if you interchange $a_k$ and $a_{k+1}$ then only the $k$-th term on the left side changes and the right side doesn't change at all.  From this we can easily see that the worst order for a given set of values is $a_1\le a_2\le\cdots \le a_n$.

Now consider adjusting both $a_k$ and $a_{k+1}$ so that the right side remains unchanged. If $a_k\mapsto a_k+\epsilon$ then $a_{k+1}\mapsto a_{k+1} - \epsilon(a_{k+1}/a_{k})^2+O(\epsilon^2)$.  If we are sitting at the worst left side for given right side, then the effect of this adjustment must be $O(\epsilon^2)$ for all $k$.  I think that this will give the worst sequence exactly by working from the last term backwards, but I'm out of time just now.

An extension of this approach is to consider adjusting three consecutive terms so that their sum and reciprocal sum both remain the same. Then only two terms of the left side will change so the total effect on the left side can be determined to obtain an explicit condition satisfied by the worst sequence.

I'll start on the last suggestion. First note that for $a,b,c\ge 0$, if we tweak them like this:
$$ a\mapsto a+\epsilon, \quad
   b\mapsto b-\frac{b^2(c^2-a^2)}{a^2(c^2-b^2)}\epsilon, \quad
   c\mapsto c+\frac{c^2(b^2-a^2)}{a^2(c^2-b^2)}\epsilon $$
then both $x+y+z$ and $1/x+1/y+1/z$ change by only $O(\epsilon^2)$.
So now tweak $a_{k-2},a_{k-1},a_k$ like that.  Up to $O(\epsilon^2)$,
the right side doesn't change and the left side changes by $\Delta_k\epsilon$,
where
$$ \Delta_k = -\frac{k-2}{S_{k-2}^2} + \frac{k-1}{S_{k-1}^2}\,
   \frac{a_k^2(a_{k-1}^2-a_{k-2}^2))}{a_{k-2}^2(a_k^2-a_{k-1}^2))}, $$
writing $S_t=\sum_{i=1}^t a_i$.  If we are looking at the largest possible
left side for given right side, then we must have $\Delta_k=0$ for $3\le k\le n$.
This determines $a_3,\ldots,a_n$ in terms of $a_1,a_2$.  We can also
take $a_1=1$ wlog, so there is one free parameter.

Note that for given $a_2$ the sequence is independent of $n$, however for large enough $n$ and any $a_2>1$, the sequence eventually becomes complex. The best value of $a_2$ depends on $n$.

Here are numerical values for the maximum value of the left side divided by $\sum_{k=1}^n 1/a_k$, and the approximate value $\hat a_2$ of $a_2$ that achieves it.

  n=2, max=1.125, $\hat a_2$=3.000  
  n=3, max=1.205, $\hat a_2$=2.426  
  n=4, max=1.261, $\hat a_2$=2.20  
  n=5, max=1.303, $\hat a_2$=2.07  
  n=10, max=1.425 $\hat a_2$=1.83