I'm trying to prove that the function
$f(t)=t\ln(1+\frac{k}{t})$; with $k\geq 0$
is increasing for $t>0$. I calculated the derivative and obtained
$f'(t)=\ln(1+\frac{k}{t})-\frac{k}{t+k}$,
but I don't know how to conclude that $f'(t)\geq0$ for $t>0$. Can someone help me, please?
In THIS ANSWER I showed using only the limit definition of the exponential function and Bernoulli's Inequality that the logarithm function satisfies the inequality
$$\bbox[5px,border:2px solid #C0A000]{\log(x)\ge \frac{x-1}{x}}\tag 1$$
for $x>0$.
Applying $(1)$, we see that
$$\begin{align} f'(t)&=\log\left(1+\frac kt\right)-\frac{k}{t+k}\\\\ &\ge \frac{k}{t+k}-\frac{k}{t+k}\\\\ &=0\end{align}$$
as was to be shown!
Find $f''(t)$
$$f''(t) = -\dfrac{k^2}{t(k+t)^2}$$
We see, $f''t < 0$ for $t > 0$. So $f'(t)$ is decreasing for $t > 0$
Also, the limit of $t \to \infty$ of $f'(t)$ is $0$:
$$\lim_{t\to\infty}f'(t) = \ln1 = 0$$
So, $f'(t)$ cannot be $< 0$ and must be $> 0$ for any finite $x$.
\begin{align} f'(t)&=\ln\left(1+\frac{k}{t}\right)-\frac{k}{t+k}\\\\ &=\int_1^{1+\frac{k}{t}}\frac{1}{x}\;dx- \frac{k}{t+k}\\\\ &\geq \int_1^{1+\frac{k}{t}}\frac{1}{1+\frac{k}{t}}\;dx- \frac{k}{t+k}\\\\ &=\frac{t}{t+k}\frac{k}{t}-\frac{k}{t+k}\\\\ &=0 \end{align}
