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MathGroup Archive 2005

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Re: How prove an inequality?

  • To: mathgroup at smc.vnet.net
  • Subject: [mg62761] Re: [mg62758] How prove an inequality?
  • From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
  • Date: Mon, 5 Dec 2005 03:37:00 -0500 (EST)
  • References: <200512041057.FAA23146@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

On 4 Dec 2005, at 19:57, Murray Eisenberg wrote:

> It's easy to argue that:  Log[2] > 1/2
>
> One argument goes as follows:  Consider
>
>     f[x_] := Log[x]
>
> on the interval from 1 to 2.  Then by the Mean Value Theorem,
> (f[2] - f[1])/(2 - 1) = f'[c] for some c with 1 < c < 2.
>
> On the one hand:
>
>     (f[2] - f[1])/(2 - 1)
> Log[2]
>
> On the other hand,
>
>     f'[c]
> 1/c
>
> and obviously 1/c > 1/2 when 1 < c < 2.
>
> Is there some simple and direct way to express the "on the other hand"
> part in Mathematica?
>
> Of course one could evaluate:
>
>    Minimize[{f'[c], 1 <= c <= 2} , c]
>
> But that does not quite express what was said above.  Surely there's
> some way to do it using Simplify or Resolve or some such function.   
> How?
>
> -- 
> Murray Eisenberg                     murray at math.umass.edu
> Mathematics & Statistics Dept.
> Lederle Graduate Research Tower      phone 413 549-1020 (H)
> University of Massachusetts                413 545-2859 (W)
> 710 North Pleasant Street            fax   413 545-1801
> Amherst, MA 01003-9305
>


How about this:

f[x_] := Log[x]


Resolve[ForAll[c,1<c<2,f'[c]>1/2]]


True

Andrzej Kozlowski


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