limsup of a weighted average Let $(a_n),(b_n),(c_n) \subset (0,+\infty)$ be three positive sequences.
Suppose $$\limsup_{N\to \infty} \frac{\sum_{n=1}^N a_n c_n}{\sum_{n=1}^N a_n} >0,\text{and}
\lim_{N\to \infty} \frac{\sum_{n=1}^N b_n c_n}{\sum_{n=1}^N b_n} =1.$$
Let $d_n = \min(a_n,b_n)$. I wonder if we have
$$\limsup_{N\to \infty}= \frac{\sum_{n=1}^N d_n c_n}{\sum_{n=1}^N d_n} >0.$$

$\limsup_{n\to \infty} A_n >0$ basically means there exists a subsequnce $A_{n_k}$ whose limit is positive. I strongly suspect that the conclusion is true after trying many examples but don't know how to prove it.

The assumption that the series are positive is necessary:
$$a_n: 0,1,1,0,0,0,0,0...$$
$$b_n: 1,0,1,0,0,0,0,0...$$
$$c_n: 2,1,0,0,0,0,0,0...$$
$$d_n: 0,0,1,0,0,0,0,0...$$
give a counterexample otherwise. So if you wanna prove this, please make sure you used this assumption in some way.
 A: It's true. Let $A_N := \{n \le N : a_n \ge b_n\}$ and $B_N := \{n \le N : b_n > a_n\}$. Then we care about $$\limsup_{N \to \infty} \frac{\sum_{n \in A_N} a_nc_n+\sum_{n \in B_N} b_nc_n}{\sum_{n \in A_N} a_n + \sum_{n \in B_N} b_n}.$$ Fix $N \ge 1$. There are two cases. The first is $\sum_{n \in A_N} a_n \le \sum_{n \in B_N} b_n$. In this case, we have $$\frac{\sum_{n \in A_N} a_nc_n+\sum_{n \in B_N} b_nc_n}{\sum_{n \in A_N} a_n + \sum_{n \in B_N} b_n} \ge \frac{1}{2}\frac{\sum_{n \in A_N} a_nc_n+\sum_{n \in B_N} b_nc_n}{\sum_{n \in B_N} b_n},$$ which is clearly lower bounded by $$\frac{1}{2}\frac{\sum_{n \in A_N} b_n c_n +\sum_{n \in B_N} b_n c_n}{\sum_{n \le N} b_n} = \frac{1}{2}\frac{\sum_{n \le N} b_nc_n}{\sum_{n \le N} b_n}.$$ In the other case (that $\sum_{n \in A_N} a_n > \sum_{n \in B_N} b_n$), we have $$\frac{\sum_{n \in A_N} a_nc_n+\sum_{n \in B_N} b_nc_n}{\sum_{n \in A_N} a_n + \sum_{n \in B_N} b_n} \ge \frac{1}{2}\frac{\sum_{n \in A_N} a_nc_n+\sum_{n \in B_N} b_nc_n}{\sum_{n \in A_N} a_n},$$ which, as the same reason as above, is at least $\frac{1}{2}\frac{\sum_{n \le N} a_n c_n}{\sum_{n \le N} a_n}$. Therefore, for each $N \ge 1$, we have $$\frac{\sum_{n \le N} d_n c_n}{\sum_{n \le N} d_n} \ge \min\left(\frac{1}{2}\frac{\sum_{n \le N} b_nc_n}{\sum_{n \le N} b_n},\frac{1}{2}\frac{\sum_{n \le N} a_nc_n}{\sum_{n \le N} a_n}\right).$$ The proof is complete by the assumptions on $(a_n)_n,(b_n)_n$, and $(c_n)_n$.
