Inequality for logarithms I conjecture the following inequality is true
$$\ln x \le (x - 1)\ln\frac{x}{x-1}$$
for all $x > 1$, but I cannot give a proof.
I will appreciate if someone can provide one. 
 A: A counter example will disprove your conjecture, take $x=3$. You'll see after suitable juggling, that, $$e \leq \dfrac{3}{4}$$ which is a serious contradiction. 
In my comment, I posted a link where Lord Wolfram computed in a considerably less time. Now, he has take pain in the form of more time to plot this inequality and show you its solution.
I thought I will also include, comment from @Bircan. He says, the reverse inequality $$\ln x \geq (x-1)\cdot\ln\left(\frac{x}{x-1}\right)$$ is true for $x \geq 2$. This additionally tells you equality holds at $x=2$, as already pointed out by Lord Wolfram in his graph.
Thank you Bircan, for you comment. 
A: Rearranging, 
$$\dfrac{1}{\ln\left(\dfrac{x}{x-1}\right)} < \dfrac{x-1}{\ln(x)}$$
the l.h.s. is asymptotic to $x$ and the r.h.s. is asymptotic to $\pi(x)$ by the PNT. But $\pi(x)/x\sim 0$. So the left hand side is growing much faster than the right hand side.(*)
(*)The key to the left hand side is the identity: 
$$\ln\left(\frac{x}{x-1}\right)= \frac{1}{x}+\frac{1}{2x^2}+\frac{1}{3x^3}+\cdots\mbox{ etc.}$$
For large $x$ this is essentially $\frac{1}{x}$, i.e.,$\frac{1}{\ln\left(\frac{x}{x-1}\right)} \sim x.$
A: The inequality is equivalent to 
$$-\frac 1x\ln \frac 1x\leq\left(1-\frac 1x\right)\ln\left(\frac 1{1-\frac 1x}\right),$$
i.e., putting $t=x^{-1}$, 
$$t\ln t\geq (1-t)\ln (1-t).$$
Put $h(t)=t\ln t$ and $g(t)=h(t)-h(1-t)=t\ln t-(1-t)\ln(1-t)$. $g'(t)=1+\ln t+h'(1-t)=1+\ln t-(-\ln(1-t)-\frac{1-t}{1-t})=2+\ln (t(1-t)),$
so $g$ is increasing in $\left(\frac 12-e^{-1},\frac 12+e^{-1}\right)$, decreasing outside this interval, and $g(\frac 12)=0$. Since $\lim_{x\to 0}g(x)=\lim_{x\to 1}g(x)=0$, we get that $g$ is non negative on $\left(\frac 12,1\right)$. It can be observed on this graphic:

