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.
 A: Fix $y > 0$, and $f(a) = \ln(a+y) - \ln y - \dfrac{a}{a+y}, a > 0\implies f'(a) = \dfrac{1}{a+y} - \dfrac{y}{(a+y)^2} = \dfrac{a}{(a+y)^2} > 0 \implies f(a) > f(0) = 0$, and the problem solved.
A: This boils down to showing that
$$f(x)=\frac{\ln(1+x)}{x}$$
is decreasing for $x>0$. Write $t=\ln(1+x)$, then
$$\frac{\ln(1+x)}{x}=\frac t{e^t-1}.$$
Does this decrease as $x$ (and so also $t$) increases? Its reciprocal,
$(e^t-1)/t$, has positive Maclaurin coefficients....
A: Use the inequality $$e^t \gt 1 + t \tag{1}$$ for $t \ne 0$
Now if $$t = -\frac{a}{a+y}$$ then $$1 + t = \frac{y}{a+y}$$
Set $t = -\dfrac{a}{a+y}$ in the above inequality (1) and take logs. That gives us
$$-\frac{a}{a+y} \gt \log (\frac{y}{a+y})$$
Move the $\log$ term to the left and use $-\log x = \log\frac{1}{x}$
A: set $t=\frac{a}{y} \gt 0$ so
$$
\log \frac{a+y}{y}-\frac{a}{a+y} = \log(1+t) +\frac1{1+t}-1 =f(t),\text{say}
$$
evidently $f(0)=0$, and
$$
f'(t)=\frac1{1+t}-\frac1{(1+t)^2} \gt 0
$$
hence $f(t)\gt 0$
A: 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 $z \gt 1,$
$\Rightarrow \star)$.
