How to prove that $\frac{\log(1+a/x)}{\log(1+1/x)}$ is increasing, for $a\geq 1$? I need to prove that the sequence 
$$
\frac{\log\big(1+\frac{a}{k}\big)}{\log\big(1+\frac{1}{k}\big)}
$$
is increasing for any $a\geq 1$, so that I thought of defining
$$ 
f(x)=\frac{\log\big(1+\frac{a}{x}\big)}{\log\big(1+\frac{1}{x}\big)}
$$
and prove that $f'(x)\geq 0$ for all $x\geq 1$. In view of some graphs I plotted, this function seems to be increasing. Moreover, it is easily seen from the expression that its limit will be precisely $a$. However, when trying to prove that $f'\geq 0$, I arrive to
$$
\frac{\log\big(1+\frac{a}{x}\big)}{1+\frac{1}{x}}-\frac{a\log\big(1+\frac{1}{x}\big)}{1+\frac{a}{x}} \overbrace{\geq}^?0.
$$
At that point I do not know how to proceed, so any hint would be kindly appreciated.
 A: For later use, consider the function $\phi(x) = x \log x$, for which $\phi'(x) = 1 + \log x$ and $\phi''(x) = \frac{1}{x}$; in particular, $\phi$ is convex on $[1, \infty)$.
If
$$
f(x) = \frac{\log\bigl(1 + \frac{a}{x}\bigr)}{\log\bigl(1 + \frac{1}{x}\bigr)},
$$
then
$$
f'(x) = \frac{(x + a) \log\bigl(1 + \frac{a}{x}\bigr) - a(x + 1) \log\bigl(1 + \frac{1}{x}\bigr)}{(x + a)(x + 1)x \log^{2}\bigl(1 + \frac{1}{x}\bigr)}.
$$
The denominator is positive for $x \geq 1$, and the numerator is equal to
\begin{align*}
(x + a) &\log\bigl(1 + \tfrac{a}{x}\bigr) - a(x + 1) \log\bigl(1 + \tfrac{1}{x}\bigr) \\
  &= (x + a) \bigl[\log(x + a) - \log x\bigr] - a(x + 1) \bigl[\log(x + 1) - \log x\bigr] \\
  &= (x + a) \log(x + a) - a(x + 1) \log(x + 1) - \bigl[(x + a) - a(x + 1)\bigr]\log x \\
  &= (x + a) \log(x + a) - a(x + 1) \log(x + 1) + (a - 1) x \log x \\
  &= \phi(x + a) - a \phi(x + 1) + (a - 1) \phi(x).
\tag{1}
\end{align*}
Since $x + 1 = \frac{a - 1}{a} x + \frac{1}{a} (x + a)$, convexity of $\phi$ implies
$$
\phi(x + 1) \leq \tfrac{a - 1}{a} \phi(x) + \tfrac{1}{a} \phi(x + a),
$$
or
$$
a \phi(x + 1) \leq \phi(x + a) + (a - 1) \phi(x).
$$
That is, (1) is non-negative, so $f$ is non-decreasing (and strictly increasing if $a > 1$, since $\phi$ is strictly convex).
