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!
 A: 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}$$
A: The inequality holds true for every $\,c\geqslant0\,$ and $\,a>b>0$.
Let $\,x=\frac ab\,$ and $\,u=\frac cb$, then $x>1$ and $u\geqslant 0$. We apply Jensen's inequality with weights in the concave case:
$$\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)}$$
and we have equality for $\,u=0\,$ only, due to the fact that $\,\log\,$ is strictly concave.
A: Assuming $a\gt b\gt0$ along with $c\gt0$, we can rewrite the inequality to prove as
$$(b+c)\log\left(a+c\over b+c\right)\ge b\log\left(a\over b\right)$$
Letting $a-b=d\gt0$, we can rewrite this further as
$$(b+c)\log\left(1+{d\over b+c}\right)\ge b\log\left(1+{d\over b}\right)$$
so what we need to prove is that the function $f(x)=x\log\left(1+{d\over x}\right)=x(\log(x+d)-\log x)$ is non-decreasing for $x\gt0$.  I.e., we need to prove that $f'(x)\ge0$ for $x\gt0$.
We see that
$$\begin{align}
f'(x)&=\log(x+d)-\log x+x\left({1\over x+d}-{1\over x}\right)\\
&=\log(x+d)-\log x-{d\over x+d}
\end{align}$$
But note that
$$\log(x+d)-\log x=\int_x^{x+d}{dt\over t}\ge\int_x^{x+d}{dt\over x+d}={(x+d)-x\over x+d}={d\over x+d}$$
and thus $f'(x)\ge0$ for $x\gt0$.
