# Proving this inequality without calculus [closed]

For all $$x\in(0,1)$$, prove that $$\ln 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'AurizioSep 24 '18 at 19:10

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• 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. – gimusi Sep 23 '18 at 17:27
• which definition of logarithm do you use? – Surb Sep 23 '18 at 20:17
• 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 $\log x$. – Paramanand Singh Sep 24 '18 at 13:54
• 1) Please avoid chamaleon questions. 2) How is $\log$ defined without Calculus? – 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 $$\log x<0 follows. (log is base $$e$$, as always).

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

We have

$$\ln (1+x)

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

$$1+x

which is true. Refer to that proof.

• Why does the last inquality hold? It fails for $x=\frac12$, for example. – user10354138 Sep 23 '18 at 17:05
• @user10354138 Yes you are right, Bernoulli indeed does not suffice! I’ve fixed that. Thanks to have pointed that out. – gimusi Sep 23 '18 at 17:17
• 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. – Misha Lavrov Sep 24 '18 at 4:15
• In the original problem it was requested without calculus. – gimusi Sep 24 '18 at 5:13