For all $x\in(0,1)$, prove that $\ln x<x$.

My attempt:

First step: Assume a function $f(x)=\ln x-x$. Analyzing tells me the function is continuous and non-differentiable at $x=0$.

Second step: $F'(x) =\frac{1}{x}-1=\frac{1-x}{x}.$

This gives me $x=1$ as an extremum. From further analysis, this appears to be a global maximum.

So combining all the info, $x=1$ is the global maximum with the function decreasing at points less than it. That would mean $f(x)$ is decreasing from $(0,1]$ and $(0,1)$. Hence, this proves the inequality.

However, is there a quick non-calculus way to solve this? I thought of using Taylor series but it's not working.


closed as off-topic by Jack D'Aurizio Sep 24 '18 at 19:10

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  • 3
    $\begingroup$ Why did you change the original problem. That’s one is trivial. You should rollback to the first one and in case create a new OP for that. $\endgroup$ – gimusi Sep 23 '18 at 17:27
  • 2
    $\begingroup$ which definition of logarithm do you use? $\endgroup$ – Surb Sep 23 '18 at 20:17
  • $\begingroup$ The fundamental inequality satisfied by logarithm is $\log x\leq x-1,\forall x>0$ and equality occurs only at $x=1$. Moreover this is an immediate consequence of any chosen definition of $\log x$. $\endgroup$ – Paramanand Singh Sep 24 '18 at 13:54
  • 2
    $\begingroup$ 1) Please avoid chamaleon questions. 2) How is $\log$ defined without Calculus? $\endgroup$ – Jack D'Aurizio Sep 24 '18 at 19:09

For $x\in (0,1)$, $\log x$ is negative (because $e^t\geq 1$ for $t\geq 0$) and so $\log x<0<x$ follows. (log is base $e$, as always).


Answer to the original problem

For all $x\in(0, 1)$ prove that $\ln(1+x) < x$ without calculus

We have

$$\ln (1+x)<x\iff 1+x<e^x$$

and by $x=\frac1y$ with $y>1$

$$1+x<e^x\iff 1+\frac1y<e^{1/y}\iff \left(1+\frac1y\right)^y<e$$

which is true. Refer to that proof.

  • 1
    $\begingroup$ Why does the last inquality hold? It fails for $x=\frac12$, for example. $\endgroup$ – user10354138 Sep 23 '18 at 17:05
  • $\begingroup$ @user10354138 Yes you are right, Bernoulli indeed does not suffice! I’ve fixed that. Thanks to have pointed that out. $\endgroup$ – gimusi Sep 23 '18 at 17:17
  • $\begingroup$ We can also prove $1 + x < e^x$ by using the Taylor series $e^x = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \dotsb$, and observing that all terms past $1+x$ are positive. $\endgroup$ – Misha Lavrov Sep 24 '18 at 4:15
  • $\begingroup$ In the original problem it was requested without calculus. $\endgroup$ – gimusi Sep 24 '18 at 5:13

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