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I need to prove that $\ln x\leq x-1 \forall x>0$, using the Mean value theorem.

For $x=1$, the equation is true.

So, for starters I'll check for $x>1$.

By applying the aforementioned theorem for $$f(t)=\ln t / [1,x]$$ we know that there is a $c\in(1,x)$ with

$$f'(c)=\frac{f(x)-f(1)}{x-1}\Leftrightarrow$$ $$\ln c=\frac{\ln x}{x-1}$$

And here I am stuck.

I know that $c>1$ thus $\ln c >\ln 1$ thus $\ln c>0$

But I don't know how to use that to prove what I need to prove.

Shevliaskovic
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  • $$1-\frac 1 x=\int_1^x t^{-2}dt\leqslant \int_1^x t^{-1}dt\leqslant \int_1^x 1 dt=x-1$$ when $x\geqslant 1$. To reverse things, take $0<y<1$ and set $x=y^{-1}>0$. You'll get the very same inequality. – Pedro Dec 12 '14 at 23:30
  • You should have $\dfrac 1 c = \dfrac{\ln x}{x-1}$, since $f'(c)=1/c$. Then use the fact that $1/c<1$ if $c>1$ and $1/c>1$ if $0<c<1$. ${}\qquad{}$ – Michael Hardy Dec 13 '14 at 00:58
  • I've posted an answer below expanding on my comment above. – Michael Hardy Dec 13 '14 at 01:07

4 Answers4

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First let's do the case where $x>1$: \begin{align} & & f'(c) & = \frac{\ln x - \ln 1}{x-1} = \frac{\ln x}{x-1} \\[10pt] \Longleftrightarrow & & \frac 1 c & = \frac{\ln x}{x-1} \\[10pt] \Longleftrightarrow & & (\text{something less than 1}) & = \frac{\ln x}{x-1} \\[10pt] \Longleftrightarrow & & (\text{something less than 1})\cdot(x-1) & = \ln x \\[10pt] \Longleftrightarrow & & \ln x & < x-1 \end{align}

The case in which $0<x<1$ is done similarly.

Michael Hardy
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  • It's hard to see why the repetition of "something less than ..." adds to the argument, why not just use $(x-1)/c$, then say "and $c>1$ in the final line. – Suzu Hirose Dec 13 '14 at 01:33
  • Because the only thing about $c$ that matters is that $1/c$ is less than $1$. ${}\qquad{}$ – Michael Hardy Dec 13 '14 at 04:50
  • You're missing some quantifiers for your propositions to be really equivalent (the term $c$ isn't defined): $\exists c\in(1,x)$ in the first two propositions, the third then reads as $0<\dfrac{\ln x}{x-1}<1$ and we're done. – gniourf_gniourf Dec 20 '14 at 14:21
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Hint: as $\log' x = 1/x$: $$ \log x - \log 1 = \frac{x-1}{c_x} $$

for some $c_x\in (1,x)$.

mookid
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Define

$$f(x):=\log x-x+1\implies f'(x)=\frac1x-1=0\iff x=1$$

and since

$$f''(x)=-\frac1{x^2}<0\;\;\forall\,x\implies x=1\;\;\text{is a maximum point} \implies$$

$$\forall\,x>0\;,\;\;f(x)\le f(1)=0$$

which is the wanted inequality.

Timbuc
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Start with $e^x \ge 1+x$, take the log (so $x \ge \ln(1+x)$), and replace $x$ with $x-1$.

I know this doesn't use the MVT, but that's the way the cookie bounces.

marty cohen
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