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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>
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