I would am interested in upper (or lower) bounds of the following ratio:
$$\frac{\sum_{j=1}^k a_j^2}{\sum_{j=1}^k a_j b_j}$$
where $a_j$ are non-negative integers, and $b_j$ are non-negative reals. As far as the numerator, I could bound using some simple options like $$\sum_{j=1}^k a_j^2\leq(\max_j a_j)^2 \cdot k$$ or via Holder $$\sum_{j=1}^k a_j^2\leq(\max_j a_j)\sum_{j=1}^k a_j$$ or even some rudimentary bound like if I assume $\sum_{j=1}^k a_j=C$, then $$\sum_{j=1}^k a_j^2\leq C\sqrt{C}.$$
Is there any obvious bound that I am over looking? If we further assume that $\sum_{j=1}^k b_j=L$, then we could write $$\begin{align} \frac{\sum_{j=1}^k a_j^2}{\sum_{j=1}^k a_j b_j}\leq&\frac{C\sqrt{C}}{\sum_{j=1}^k a_j b_j} \\ \leq& \frac{C\sqrt{C}}{\sqrt{\sum_{j=1}^k a_j^2} \sqrt{\sum_{j=1}^k b_j^2}}\\ \leq& \frac{C\sqrt{C}}{\sqrt{C\sqrt{C}}\sqrt{L\sqrt{L}}}=\frac{C^\frac{3}{4}}{L^\frac{3}{2}} \end{align}$$
Consider $A=(a_1,a_2,\cdots a_n)$ and $B=(b_1,b_2,\cdots b_n)$ as $n$ dimensional vectors.
Your expression can be written as the square of a norm divided by a dot product:
$$R=\|A\|^2/(A \cdot B)=\|A\|^2/(\|A\|\|B\| \cos \alpha)=\|A\|/(\|B\| \cos \alpha)$$
where the cosine is positive due to the positivity of the coordinates.
Therefore the minimal value of $R$ is
$$\|A\|/\|B\|=\sqrt{\sum a_i^2/\sum b_i^2 }$$
knowing that its maximal value can be arbitrarily large.
Remark: the fact that the $a_i$ are integers and the $b_i$ are real doesn't play any role.