What is the simplest way to show that $x\log(1+\frac{a}{x})$ is an increasing function of $x$? How to show that $x\log(1+\frac{a}{x})$ is an increasing function of $x$ for non-negative values of $x$? Actually its derivative is $$\log\left(1+\frac{a}{x}\right)-\frac{a}{a+x}$$ ($a$ is positive) where we can see that the second term is negative> so how to show that the derivative of $x\log(1+\frac{a}{x})$ is positive? Thanks in advance.
 A: I think the most parsimonious way to do this is via the integral representation of the logarithm. To start, note that
$$\begin{split}\frac{\ln(1+t)}{t}&=\frac{1}{t}\int_1^{1+t}\frac{\mathrm{d}t'}{t'}\\
&=\frac{1}{t}\int_0^{t}\frac{\mathrm{d}u}{1+u}\\
&=\int_0^{1}\frac{\mathrm{d}v}{1+tv}\text{.}
\end{split}$$
Substitute $t\leftarrow a/x$ and multiply both sides by $a$. Then
$$x\ln(1+a/x)=\int_0^{1}\frac{ax\mathrm{d}v}{x+av}$$
so
$$\frac{\mathrm{d}}{\mathrm{d}x}\left(x\ln(1+a/x)\right)=\int_0^{1}\frac{a^2v\mathrm{d}v}{(x+av)^2}$$
which is manifestly positive.
A: The function $g(u) = (1 + u)\log(1 + u) - u$ is increasing for $u > 0$. Indeed, 
$$
g^{\prime}(u) = \log(1 + u) + \dfrac{1 + u}{1 + u} - 1 = \log(1 + u) \geq 0 \quad \text{to} \quad u \geq 0   
$$
But, $g(0) = 0$. Thus, $g(u) > g(0) = 0$ if $u > 0$ or
$$
(1 + u)\log(1 + u) > u \quad \Rightarrow \quad \log(1 + u) - \dfrac{u}{1 + u} > 0
$$ 
for $u > 0$. For $u = a/x$, we have
$$
\log\biggl(1 + \dfrac{a}{x}\biggr) - \dfrac{a/x}{1 + a/x} > 0 \quad \Rightarrow 
$$
$$
f^{\prime}(x) = \log\biggl(1 + \dfrac{a}{x}\biggr) - \dfrac{a}{a + x} > 0 \quad \Rightarrow
$$
$f$ increasing.
A: Using the integral definition of $\log$,
$\begin{array}\\
(x\log(1+\dfrac{a}{x}))'
&=\log\left(1+\frac{a}{x}\right)-\dfrac{a}{a+x}\\
&=\int_1^{1+\frac{a}{x}}\dfrac{dt}{t}-\dfrac{a}{a+x}\\
&=\int_0^{\frac{a}{x}}\dfrac{dt}{1+t}-\dfrac{a}{a+x}\\
&=\int_0^{\frac{a}{x}}\dfrac{dt}{1+t}-\dfrac{a/x}{1+a/x}\\
&=\int_0^{\frac{a}{x}}\left(\dfrac{1}{1+t}-\dfrac{1}{1+a/x}\right)dt\\
&=\int_0^{\frac{a}{x}}\left(\dfrac{(1+a/x)-(1+t)}{(1+t)(1+a/x)}\right)dt\\
&=\int_0^{\frac{a}{x}}\left(\dfrac{a/x-t}{(1+t)(1+a/x)}\right)dt\\
&\ge 0
\qquad\text{since } \frac{a}{x} \ge t\\
\end{array}
$
