Is there any other way to get upper bound of the ration of the weighted summation? Fix some constants $c_1, c_2,\dots, c_n$ so that $\sum_{i=1}^n c_i^2=1$ and $|c_i|\le 1$ for $i=1,\dots, n$, and $-2\le b_1\le \dots \le b_n\le 2$.
I want to upper bound the following equation
$$
L=\frac{c_1^2e^{-4b_1t}}{\sum_{i=1}^n c_i^2 e^{-4b_it}}.
$$
It is clear that $L\le 1$.
(1) One trivial upper bound is that since $\sum_{i=1}^n c_i^2 e^{-4b_it}\ge c_2^2 e^{-4b_2t}$, so we have that
$$
L\le \frac{c_1^2e^{-4b_1t}}{c_2^2e^{-4b_2t}}\le \frac{e^{-4b_1t}}{c_2^2e^{-4b_2t}}
$$
But I don't want to have a proportionality constant like $c_1^2/c_2^2$ or $1/c_2^2$ for some reason.
Question: Is there any other way to get its upper bound? The coefficients depend only on $c_1$ or $c_i$, not on the ratio of constants $c_i$.
 A: We solve a more general problem by finding the upper bound of
$$L=\frac{c_\color{blue}{k}^2e^{-4b_\color{blue}{k}t}}{\sum_{i=1}^n c_i^2 e^{-4b_it}} \qquad \quad \text{for }0\le k\le n$$
We have
$$\begin{align}
L &= \frac{c_k^2e^{-4b_kt}}{\sum_{i=1}^n c_i^2 e^{-4b_it}}=\frac{c_k^2}{\sum_{i=1}^n c_i^2 e^{-4(b_i-b_k)t}}=\frac{c_k^2}{\sum_{i<k} c_i^2 e^{-4(b_i-b_k)t}+c_k^2+\sum_{i>k} c_i^2 e^{-4(b_i-b_k)t}}\\
\end{align}$$
As $-2\le b_1 \le ...\le b_n \le 2$, we have : $-4 \le b_i- b_k \le 0$ for $i<k$ and $0 \le b_i- b_k \le 4$ for $i>k$, we deduce that
$$e^{-4(b_i-b_k)t}\ge 1 \qquad \text{     for }i <k$$
$$e^{-4(b_i-b_k)t}\ge e^{-16t} \qquad \text{for }i >k$$
Then
$$L \le \frac{c_k^2}{\left(\sum_{i<k} c_i^2\right) +c_k^2+\left(\sum_{i>k} c_i^2\right) e^{-16t}} =\frac{c_k^2}{\left(\sum_{i \le k} c_i^2\right)+\left(1-\left(\sum_{i \le k} c_i^2\right)\right) e^{-16t}}$$
The equality occurs when $b_1 = ...=b_k=-2$ and $b_{k+1}=...=b_n=2$.
Remark: By the same method, the lower bound is
$$L \ge \frac{c_k^2}{\left(\sum_{i < k} c_i^2\right)e^{16t} +1-\sum_{i < k} c_i^2 }$$
when $b_1 = ...=b_{k-1}=-2$ and $b_{k}=...=b_n=2$
For your initial problem,
$$\color{red}{L \le \frac{c_1^2}{c_1^2+\left(1-c_1^2\right) e^{-16t}}}$$
