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