Inequality involving sum of exponentials Let
$\lambda = \frac{\sum\limits_{n=1}^{\infty} \lambda_n B_n e^{-\lambda_n t_0}}{\sum\limits_{n=1}^{\infty} B_n e^{-\lambda_n t_0}}\,,$
with $\lambda_n$, $B_n$, $t_0$ positive real numbers.
Does the following always hold for $t_1 > t_0$
$\frac{\sum\limits_{n=1}^{\infty}  B_n e^{-\lambda_n t_1}}{\sum\limits_{n=1}^{\infty} B_n e^{-\lambda_n t_0}} \ge 
\frac{e^{-\lambda t_1}}{e^{-\lambda t_0}}\,?
$
[Edit: inequality sign corrected from '<' to $\ge$ after Kavi Rama Murthy's answer and random's comment.]
 A: You cannot have such an inequality. RHS $<1$. If such an inequality holds for all choices of $B_n$'s we can get $1$ as  a limting value of LHS leading to a  contradiction. For example you can let $B_n$ and $\lambda_n$ approach $0$ for all $n>1$, $\lambda_1=1$ and $B_1$ approach $\infty$. 
A: Yes, let $f(t)=\sum\limits_{n=1}^{\infty} B_n e^{-\lambda_n t}$ and note that $\sum\limits_{n=1}^{\infty} \lambda_n B_n e^{-\lambda_n t}=-\sum\limits_{n=1}^{\infty} \frac d{dt}(B_n e^{-\lambda_n t})$.
The problem statement implies that both series converge for $t=t_0$ and therefore uniformly converge for $t\ge t_0$, so $\sum\limits_{n=1}^{\infty} \frac d{dt}(B_n e^{-\lambda_n t})=\frac d{dt}\sum\limits_{n=1}^{\infty} (B_n e^{-\lambda_n t})$ and one has $$\lambda(t) = \frac{\sum\limits_{n=1}^{\infty} \lambda_n B_n e^{-\lambda_n t}}{\sum\limits_{n=1}^{\infty} B_n e^{-\lambda_n t}}=-\frac{d \log f}{dt}$$
The expression for $\lambda(t)$ can also be interpreted as a weighted arithmetic mean of the individual $\lambda_n$ and since with growing $t$ the normalized weights for smaller $\lambda_n$ increase at the expense of those of larger $\lambda_n$, $\lambda$ must be a non-increasing function of $t$.
By the mean value theorem there is a $t_i \in (t_0,t)$ for which $\log f(t)-\log f(t_0)=-\lambda(t_i)(t-t_0)\ge -\lambda(t_0)(t-t_0)$, which is equivalent with $\frac{f(t)}{f(t_0)}\ge e^{-\lambda(t_0)(t-t_0)}$.
