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I need to prove that $$\frac {x-1}{x} \leq \ln x $$ using only logarithmic properties and the fact that $x-1\geq \ln x$

I've been twisting and turning the inequality for a while now. I tried this; starting from what we're trying to prove, and working backwards:

$$\frac {x-1}{x} \leq \ln x \Leftrightarrow x-1 \leq x\ln x $$

We know that $x-1\geq \ln x$ which implies: $\ln x \leq x\ln x$ which is true for all the $x$s allowed.

Am I doing this completely wrong? Could I have a hint to push me in the right direction.

(I have of course tried to twist and turn the inequality using logarithmic properties, but I keep walking in circles…)

egreg
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user9750060
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    Hint : Use the property $x-1\ge ln(x)$ for the number $\frac{1}{x}$ – Peter Sep 28 '17 at 20:21
  • Would you mind editing your comment? Is it supposed to say ln(x)? And I tried that method but didn't get anywhere. But I'll try again, thanks. – user9750060 Sep 28 '17 at 20:23

1 Answers1

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Set $x=1/X$ with $X>0$. What you have to prove is that:

$$\dfrac{\tfrac1X-1}{\tfrac1X} \leq ln(\dfrac1X)$$

itself equivalent to:

$$1-X \leq -ln(X)$$

Can you conclude ?

Jean Marie
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