Re: Inequality

*To*: mathgroup at smc.vnet.net*Subject*: [mg122023] Re: [mg121999] Inequality*From*: Andrzej Kozlowski <akoz at mimuw.edu.pl>*Date*: Sun, 9 Oct 2011 03:53:17 -0400 (EDT)*Delivered-to*: l-mathgroup@mail-archive0.wolfram.com*References*: <201110080935.FAA21078@smc.vnet.net>

On 8 Oct 2011, at 11:35, ff mm wrote: > (a + b)/2 > (b - a)/(Log[b] - Log[a]) > Sqrt[a b] && a > 0 && b > o && > a != b > How to judge the inequality is correct or not ? > > When I used "Refine" , I got the results I didn't want . > > In[1]:= Refine[(a+b)/2>(b-a)/(Log[b]-Log[a])>Sqrt[a b],a>0&&b>o&&a! > =b] > Out[1]=(a + b)/2 > (-a + b)/(-Log[a] + Log[b]) > Sqrt[a] Sqrt[b] > > I want to get the "True", how ? > > Thanks > Sincerely Adam > I think you need a little mathematics to prove this. Let us start with first condition (a + b)/2 > (b - a)/(Log[b] - Log[a]) Under the given assumptions this can be written as (a+b)/2 > (b-a)/Log[b/a] dividing both sides by a we get (1+z)/2 >(z-1)/Log[z] where z = b/a. Let's now ask Mathematica when this is true: In[166]:= Reduce[(1+z)/2>(z-1)/Log[z],z] Out[166]= 0<z<1||z>1 Hence it is true precisely when a!=b and both a and b are positive, as you wanted. Now let's do the same with the other condition (b - a)/(Log[b] - Log[a]) > Sqrt[a b] i.e. (b-a)/Log[b/a]>Sqrt[a b] dividing by a we get (z-1)/Log[z]>Sqrt[z] Reduce[(z - 1)/Log[z] > Sqrt[z], z] 0 < z < 1 || z > 1 Same as before and we are done. Reduce is a fantastic but I think it can only do such marvels with a single variable. Andrzej Kozlowski

**References**:**Inequality***From:*ff mm <venture.knowledge@gmail.com>