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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.

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Lord Wolfram says, the solution of your inequality is $1<x \leq 2$, and not $x > 1$. So, your conjecture is disproved!! wolframalpha.com/input/… –  user21436 Dec 24 '11 at 16:47
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Actually, for $x \geq 2$, ln$x \geq (x-1)$ln$(x/x-1)$ is true. wolframalpha.com/input/… –  James Dec 24 '11 at 16:52
    
Could you explain how you conjecture this? –  Vladimir Dec 24 '11 at 16:52
    
Make the substitution $y = x - 1$ and exponentiate to get the conjectured inequality of $1+y \leq (1+1/y)^y$ for all $y > 0$. Since the right-hand side converges to $e$ as $y \to \infty$, then the conjecture is obviously false. –  cardinal Dec 24 '11 at 19:42

3 Answers 3

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.

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Thank you Kannappan. The inequality is only locally true. Too bad Lord Wolfram doesn't give us analytical proof, but I believe him. –  Martin Dec 24 '11 at 18:14

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:

enter image description here

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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.,$\frac{1}{\ln\left(\frac{x}{x-1}\right)} \sim x.$

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I would appreciate help in figuring out the second equation, i.e, the limit is 1/2. Please elaborate. –  user21436 Dec 24 '11 at 17:15
    
@Kannapppan: thanks for the edit. latex and i are not yet compatible. –  daniel Dec 24 '11 at 17:16
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@Kannappan: also Lord Wolfram! But I noticed that numerically the l.h.s. is close to x. I will hopefully be able to add a proof but will have to wait til after holidays. –  daniel Dec 24 '11 at 17:29
    
Shall definitely be looking forward to it. And, I shall also think about it till then. I actively participate in SE only because, questions such as this come up so often, that you'll learn a lot. –  user21436 Dec 24 '11 at 17:42

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