How to prove that $(x+c)\log(\frac{c+x}{x})>c$ How to prove that $(x+c)\log(\frac{c+x}{x})>c$ for $x, c > 0$? For $\frac{c+x}{x} \ge e$ it's obvious.
 A: Letting $x = cy, y > 0$, your inequality can be simplified to $(y+1)\log\left(\frac{y+1}{y}\right) > 1$, or $\log\left(\frac{y+1}{y}\right) > \frac{1}{y+1}$. This inequality then follows from substituting $t = \frac{1}{y}$ in the inequality $\log(1 + t) > \frac{t}{t+1},t > 0$ (which can be proven by the mean value theorem).
A: Hint: let $\frac{c+x}{x} = u, u>1$. What is $\frac{c}{x+c}$ in terms of $u$?
A: $(x+c)\ln\left(\dfrac{x+c}{x}\right) > c\iff f(c) =(x+c)\ln(x+c) - (x+c)\ln x - c > 0\Rightarrow f'(c) = \ln(x+c)+\dfrac{x+c}{\ln(x+c)}-\ln x - 1 > c + 0 + 0 = c > 0\Rightarrow f(c) > \displaystyle \lim_{c \to 0^{+}} f(c) = 0$, and you're done.
A: In THIS ANSWER, I showed using only the limit definition of the exponential function and Bernoulli's Inequality that the logarithm function satisfies the inequalities
$$\frac{z-1}{z}\le \log(z)\le z-1 \tag 1$$
for $z>0$.  Letting $z=\frac{c+x}{x}$ in $(1)$ gives immediately 
$$\frac{c}{c+x}\le \log\left(\frac{c+x}{x}\right)\le \frac{c}{x} \tag 2$$
whereupon multiplying $(2)$ by $c+x$, assuming $c+x>0$ yields the coveted inequality
$$\bbox[5px,border:2px solid #C0A000]{(c+x)\log\left(\frac{c+x}{x}\right)\ge c}$$
