I will add an answer here to flesh out some of the details of the construction given in the previous answer. In particular, to check thoroughly that $(c_n)_n$ is indeed decreasing.
Suppose that $(a_n)_n$ and $(b_n)_n$ are positive and non-increasing, let $(A_n)_n$ and $(B_n)_n$ be the respective partial sum sequences, and assume that $A_n$ and $B_n$ both approach infinity as $n$ approaches infinity. Observe that both $(A_n)_n$ and $(B_n)_n$ are necessarily increasing (because $(a_n)_n$ and $(b_n)_n$ are positive).
As before, let $C^*_n = \sqrt{\min(A_n, B_n)}$ for each $n$. Then the following are easily checked.
- $C^*_n/A_n \to 0$ and $C^*_n/B_n \to 0$ as $n \to \infty$ by the squeeze theorem.
- $C^*_n \to \infty$ as $n\to \infty$, because the same holds for both $(A_n)_n$ and $(B_n)_n$. In particular, for any fixed $M$ it holds that both $A_n \geq M^2$ and $B_n \geq M^2$ for all sufficiently large $n$, in which case $C^*_n \geq M$ for all sufficiently large $n$ as well.
- $(C^*_n)_n$ is increasing. Indeed, fix $n$ and suppose WLOG that $C^*_n = \sqrt{A_n}$ (in which case $A_n \leq B_n$). Then in the first case, $C^*_{n+1} = \sqrt{A_{n+1}} > \sqrt{A_n} = C^*_n$ since $(A_n)_n$ is increasing. In the other case, $C^*_{n+1} = \sqrt{B_{n+1}} > \sqrt{B_n} \geq \sqrt{A_n} = C^*_n$ since $(B_n)_n$ is increasing (and by assumption that $A_n \leq B_n$).
We might then proceed to define $c^*_1 = C^*_1$ and $c^*_n = {}^-\Delta C^*_n = C^*_n - C^*_{n-1}$ for $n > 1$. It holds that $(c^*_n)_n$ is positive by the third property above, it holds that $\sum_{n=1}^N c^*_n = C_N$ for each $N$ by telescoping series (which therefore diverges to infinity by the second property above), and it holds that the divergence is slower than that of either $(a_n)_n$ or $(b_n)_n$ in the desired sense by the first property above.
It remains to verify that $(c^*_n)_n$ is decreasing. But this is not obvious, and possibly not true in general. However, the definition can be slightly modified to ensure that this holds.
Imagine a polygonal path joining each point $(n, C^*_n)$ on a graph. In essence, we have a path in the $xy$-plane which diverges to infinity much slower than either $(n, A_n)$ or $(n, B_n)$, but the vertical increments ${}^-\Delta C_n$ may be non-monotone (i.e., the sequence of slopes of each linear segment may not be decreasing). We can ensure that the increments are monotone by translating each point further and further to the right, until the slopes of each linear segment form a decreasing sequence.
To make this precise, we shall (inductively) construct a sequence of indices $(n^*_k)_k$ and define $C_n = C^*_k$ whenever $n = n^*_k$. For those $n$ such that $n^*_k < n < n^*_{k+1}$, we shall define $C_n$ by a linear interpolation. That is,
$$ C_n = C^*_k + m_k(n - n^*_k) \quad \text{where} \quad m_k = \frac{C^*_{k+1} - C^*_k}{n^*_{k+1} - n^*_k} $$
Begin with $n^*_1 = 1$ and $n^*_2 = 2$. Then define $n^*_3$ as the smallest $n$ such that
$$ \frac{C^*_3 - C^*_2}{n - n^*_2} \leq m_1 $$
where $m_1 = (C^*_2 - C^*_1)/(n^*_2 - n^*_1)$ as defined above. Continue inductively; for each $k$, define
$$n^*_k = \min\Big\{n : \frac{C^*_k - C^*_{k-1}}{n - n^*_{k-1}} \leq m_{k-2} \Big\}$$
We note here that by isolating $n$ in the above inequality, it is possible to present $n^*_k$ in exact form as the following.
$$ n^*_k = \left\lceil n^*_{k-1} + (n^*_{k-1} - n^*_{k-2})\left(\frac{C^*_k - C^*_{k-1}}{C^*_{k-1} - C^*_{k-2}}\right) \right\rceil $$
Finally, define $(C_n)_n$ as described above. Observe that $(C_n)_n$ is still increasing, still diverges (because it is increasing and contains a divergent subsequence, namely $C_{n^*_k} = C^*_k$), and the divergence is still slower than that of either $(A_n)_n$ or $(B_n)_n$ because $C_n \leq C^*_n$. Finally, define $c_1 = C_1$ and $c_n = {}^-\Delta C_n = C_n - C_{n-1}$ for each $n > 1$. By construction, $c_{n+1} \leq c_n$ for each $n$ as desired, thus everything is satisfied.
