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!

Let $a,b,c \in \mathbb{R}^+$ and $a>b$.
Assume
$$\left(\frac{a+c}{b+c}\right)^{b+c}>\left(\frac{a}{b}\right)^{b}$$
Then
$$\log\left(\left(\frac{a+c}{b+c}\right)^{b+c}\right)>\log\left(\left(\frac{a}{b}\right)^{b}\right)$$(as both sides are positive and $\log$ is an increasing function)
$$\implies (b+c)\log\left(\frac{a+c}{b+c}\right)>b\log\left(\frac{a}{b}\right)$$
$$\implies \frac{b+c}{b}>\frac{\log\left(\frac{a}{b}\right)}{\log\left(\frac{a+c}{b+c}\right)}$$
(note for the rearrangement on the last line we have to make sure $\log\left(\frac{a+c}{b+c}\right)$ is positive which is true as $\log\left(\frac{a+c}{b+c}\right) = \log\left(a+c\right)-\log\left(b+c\right)$ and $a>b$ and $\log$ is an increasing function)

All we need to show now is that
$$\left(\frac{a+c}{b+c}\right)^{b+c}>\left(\frac{a}{b}\right)^{b}$$
I haven't been able to find an elegant way, so I've resorted to calculus (the idea being to write the LHS as $y(c)$ and the RHS as $y(0)$ for some function $y$, and then showing that $y(c)>y(0)$ by showing y is a strictly increasing function).
Let $y(x)=$$\left(\frac{a+x}{b+x}\right)^{b+x}$, $x \in \mathbb{R}^+$
Then
$$\log y=(b+x)\left(\log(a+x) - \log(b+x)\right)$$
$$\implies y'y^{-1}=(b+x)((a+x)^{-1} - (b+x)^{-1})+(\log(a+x) - \log(b+x))$$
$$\implies y'=y((b+x)(a+x)^{-1} - 1 +\log(a+x) - \log(b+x))$$
$$=y\left(\frac{(b+x) - (a+x)}{a+x} +\log(a+x) - \log(b+x)\right)$$
$$y>0 \,\,\,\forall\,x\in\mathbb{R}^+\implies y' > \frac{(b+x)-(a+x)}{a+x} +\log(a+x) - \log(b+x))$$
Let $g(x)=\log x$
Then
$$y' > ((a+x)-(b+x))(-g'(a+x)) + g(a+x)-g(b+x)$$
$$a-b>0\implies y'((a+x)-(b+x))^{-1} >\frac{g(a+x)-g(b+x)}{(a+x)-(b+x)}-g'(a+x)>0$$
Observe on the graph of $y=\log x$ that the slope of the line between 2 points ($(b+x)$ and $(a+x)$ in this case) is greater than the slope at the rightmost point. Hence $\frac{g(a+x)-g(b+x)}{(a+x)-(b+x)}>g'(a+x)$. (Alternatively, an algebraic argument can be made using the intermediate value theorem on $g'(x)$ and that $g''(x)<0\,\,\,\forall\,x\in\mathbb{R}^+$)
$$\implies y' > 0$$
Hence $y(x)$ is a strictly increasing function
$$\implies y(c)>y(0)$$
$$\implies \left(\frac{a+c}{b+c}\right)^{b+c}>\left(\frac{a}{b}\right)^{b}$$
