       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:= Refine[(a+b)/2>(b-a)/(Log[b]-Log[a])>Sqrt[a b],a>0&&b>o&&a!
> =b]
> Out=(a + b)/2 > (-a + b)/(-Log[a] + Log[b]) > Sqrt[a] Sqrt[b]
>
> I want to get the "True", how ?
>
> Thanks
>

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:= Reduce[(1+z)/2>(z-1)/Log[z],z]
Out= 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>
• Prev by Date: Re: Sort on vector component
• Next by Date: Re: mathematica antialiasing quality has no influence
• Previous by thread: Inequality
• Next by thread: Re: Inequality