Inequality: $\ln(\frac{a+y}{y})-\frac{a}{a+y}> 0$ I want to show that the function $g(y)=y\ln(1+\frac{a}{y})$ is increasing for $y>0, a>0$. 
I've found the derivative and set up the inequality that I need to show:
$\ln(\frac{a+y}{y})-\frac{a}{a+y}> 0$
I'm not sure about how to show it. Would appreciate a suggestion or hint.

Problem: 
$\ln (\frac{a+y}{y}) - \frac{a}{a+y} \gt 0$, for  $a, y, \gt 0$.
LHS:
$\ln( \frac{a+y}{y}) - 1 + \frac{y}{a+y}$.
Let $z: = \frac{a+y}{y}$ , then $z \gt 1$.
Problem reduces to:
$\star)$ $\ln (z) + \frac{1}{z} \gt 1$ for $z \gt 1$.
$f(z) := \ln(z) + \frac{1}{z}$ ;
$f(1) = 0+ 1 = 1$.
$f'(z) = \frac{1}{z} - \frac{1}{z^2} \gt 0$ for $z \gt 1$.
Thus $f(z)$ is stricly monotonically increasing for $z \gt 1,$
$\Rightarrow \star)$.
