The log sum inequality states that
$ \sum_i a_i\ln\frac{a_i}{b_i}\geq a\ln{\frac{a}{b}}$
where $a=\sum_i a_i$ and $b=\sum_i b$.
Is there a generalisation (with whatever conditions) that extends it the following situation?
$\sum_i a_i\ln\frac{b_i}{c_i}$
Not really, without knowing more about the coefficients. Following gammatester's comment,
$$a_i\ln\frac{b_i}{c_i}=a_i\ln\left(\frac{a_i}{c_i} \frac{b_i}{a_i}\right) =a_i\ln\frac{a_i}{c_i}-a_i\ln\frac{a_i}{b_i}$$
But this would not give a useful bound as we have a difference of two upper-bounded quantities. At most, you can say:
$$\sum_i a_i\ln\frac{b_i}{c_i} \ge A \ln \frac{A}{C} + \sum_i a_i\ln\frac{b_i}{a_i} $$
To bound that we'd need some bound in the other direction $\sum_i a_i\ln\frac{b_i}{a_i} \ge f(A,B)$ but that's not possible in general (think of $A=B=1$, we can do that as small as we want by letting one of the $b_i \to 0$). For the particular case $a_i/b_i=\alpha$ it's certainly possible, but (looking at the original), it's also trivial. Also, if we impose $b_i/a_i\ge \beta$, then the RHS can be bounded as $A \ln \frac{A}{C} + A \ln\beta $, but again, this does not look very interesting.