logarithm proof I'm trying to prove the following inequalities:
\begin{align}
\frac{b+c}{b} \geq \frac{\log(\frac{a}{b})}{\log(\frac{a+c}{b+c})}, c \in (0,1)
\end{align}
I know that $\log(\frac{a}{b}) > \log(\frac{a+c}{b+c})$ for $ c > 0$, but I'm stuck as to how to proceed with the proof. This is not a homework and is a part of larger proof. This inequality may turn out to be false. 
Additional constraint is that $a > b$.
Any hints will be appreciated!

The inequality holds true for every $\,c\geqslant0\,$ and $\,a>b>0$.
Let $\,x=\frac ab\,$ and $\,u=\frac cb$, then $x>1$ and $u\geqslant 0$. We apply Jensen's inequality with weights in the concave case:
$$\log\left(\frac{1}{1+u}\cdot x +\frac{u}{1+u}\cdot 1\right) \;\geqslant\;
\frac{1}{1+u}\log (x)\:+\: \frac{u}{1+u}\log(1) \\[4ex]
\implies\: 1+u\;\geqslant\;\frac{\log(x)}{\log\big(\frac{x+u}{1+u}\big)}$$
and we have equality for $\,u=0\,$ only, due to the fact that $\,\log\,$ is strictly concave.
