Inequality with natural logarithm I would like to prove the following inequality: $\forall t\in (0,1), n\in\mathbb{N}$:
$$\frac{\ln(1+nt)}{\ln(1+n)}<\frac{\ln(1+nt+t)}{\ln(1+n+1)}.$$
I defined the function $f_n(t):=\frac{\ln(1+nt)}{\ln(1+nt+1)}$ and now tried to show the inequality $f_n(t)< f_n(1)$, $\forall t\in(0,1)$, but I could not prove it yet. How can one prove the above inequality? 
Best regards
 A: Following your idea, calculate
$$
f'_n(t) = \frac{n\frac{\ln(1+(n+1)t)}{1+nt} - (n+1)\frac{\ln(1+ nt)}{1+(n+1)t}}{\bigl(\ln(1+(n+1)t)\bigr)^2}.
$$
Next we will prove that 
$$
n\frac{\ln(1+(n+1)t)}{1+nt} \gt (n+1)\frac{\ln(1+ nt)}{1+(n+1)t}, 
$$
or, equivalently,
$$
\bigl(1+(n+1)t\bigr)\frac{\ln(1+(n+1)t)}{n+1} \gt (1+nt)\frac{\ln(1+ nt)}{n}.
$$
The last inequality is a consequence of the fact that the function
$$
g(t,n) = (1+nt)\frac{\ln(1+ nt)}{n}
$$
is an increasing function of $n$ since
$$
\frac{d}{d n}g(t,n) = \frac{nt - \ln(1+ nt)}{n^2},
$$
which is clearly positive. 

This might be somewhat simpler: We will prove that for $t \in (0,1)$ the function 
$$
h(t,n) = \frac{\ln(1+nt)}{\ln(1+n)}
$$ 
is an increasing function of $n$. Calculate
$$
\frac{d}{dn} h(t,n) = \frac{\frac{t \ln(1+n)}{1+nt} - \frac{\ln(1+nt)}{1+n}}{(\ln(1+n))^2}. 
$$ 
The derivative $\frac{d}{dn} h(t,n)$ is positive if and only if 
$$
(1+nt)\ln(1+nt) \lt t(1+n) \ln(1+n). 
$$
The last inequality is true since the function $k(t,n)=(1+nt)\ln(1+nt)$ is a convex function of $t$ on $(0,1)$ and $k(0,n) = 0$ and $k(1,n) = (1+n)\ln(1+n)$. For the convexity, just calculate 
$$
\frac{d^2}{d^2t} k(t,n) = \frac{n^2}{1+nt}.
$$
