Prove that $(1+\frac{\log(x)}{x})^x$ is monotonically increasing function for $x \geq 1$

Question

Let $f(x) = (1+\frac{\log(x)}{x})^x$. How do we prove that $f(x)$ is a monotonically increasing function for $ x \geq 1$?

This question is related to How to prove $(1+1/x)^x$ is increasing when $x>0$? and Proof that $(1+1/x)^x$ is monotonic increasing

Answer 1

$$f(x)=(1+\frac{\ln(x)}{x})^x$$

$$\ln f(x)=x\ln(1+\frac{\ln(x)}{x})$$

$$\frac{f'(x)}{f(x)}=x\frac{1-\ln x}{x^2(1+\frac{\ln(x)}{x})}+\ln(1+\frac{\ln(x)}{x})$$

$$\frac{f'(x)}{f(x)}=\frac{1-\ln x}{x(1+\frac{\ln(x)}{x})}+\ln(1+\frac{\ln(x)}{x})$$

$f'(x)\geq 0$ for $ x\geq1$ obviously

So $f(x)$ is monotonic increasing for $x \geq 1$

Answer 2

Putting $x=\exp(t)$ with $t\geq 0$ give that the function is increasing if and only if $G(t)=\exp(t)\log (1+t\exp(-t))$ is increasing. Computing the derivative and multiplying by $\exp(-t)$, we have to show that, with $u=t\exp(-t)$, we have $\displaystyle \log (1+u)-\frac{u}{1+u}+\frac{\exp(-t)}{1+u}\geq 0$ for $u\geq 0$. It is sufficient to show that $\displaystyle \log (1+u)-\frac{u}{1+u}\geq 0$ for $u\geq 0$, and this is easy to do.

Answer 3

For $x\ge1$, $f(x)\gt0$, so $f(x)$ is increasing if and only if

$$g(x)=\ln f(x)=x(\ln(x+\ln x)-\ln x)$$

is increasing, so we need only prove $g'(x)\ge0$ for $x\ge1$. In fact we have

$$\begin{align} g'(x)&=(\ln(x+\ln x)-\ln x)+x\left({1+{1\over x}\over x+\ln x}-{1\over x} \right)\\ &=(\ln(x+\ln x)-\ln x)+\left({x+1\over x+\ln x}-1\right)\\ &=(\ln(x+\ln x)-\ln x)+{1-\ln x\over x+\ln x}\\ &=\int_x^{x+\ln x}{dt\over t}+{1-\ln x\over x+\ln x}\\ &\ge\int_x^{x+\ln x}{dt\over x+\ln x}+{1-\ln x\over x+\ln x}\\ &={\ln x\over x+\ln x}+{1-\ln x\over x+\ln x}\\ &={1\over x+\ln x}\gt0 \end{align}$$

The key step was using the integral definition of the natural logarithm and the decreasing nature of $1\over t$ to write

$$\ln(a+b)-\ln a=\int_a^{a+b}{dt\over t}\ge\int_a^{a+b} {dt\over a+b}={(a+b)-a\over a+b}={b\over a+b}$$