Re: Re: which one is greater than or equal?
- To: mathgroup at smc.vnet.net
- Subject: [mg41348] Re: [mg41299] Re: [mg41285] which one is greater than or equal?
- From: Bobby Treat <drmajorbob+MathGroup3528 at mailblocks.com>
- Date: Thu, 15 May 2003 04:06:18 -0400 (EDT)
- Sender: owner-wri-mathgroup at wolfram.com
Yes, and it's too obvious to use the AG inequality on it, too.
I don't think anybody in America proves this in junior high school (or
even high school, I suspect). We don't have teachers that bright here.
Anyway, Jensen's Inequality is a nicer way to deal with the original
problem, as it allows us to substitute any concave function for Sqrt
and get the same result.
Bobby
-----Original Message-----
From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
To: mathgroup at smc.vnet.net
Subject: [mg41348] Re: [mg41299] Re: [mg41285] which one is greater than or equal?
Well, honestly, I took this as too obvious to state. It is after all
the way one proves it in (junior?) high school.
Andrzej
On Thursday, May 15, 2003, at 01:33 am, Bobby Treat wrote:
> When you evaluate
>
> Expand[4*(g[2]^2 - f[2]^2)]
>
> a[1] - 2*Sqrt[a[1]]*Sqrt[a[2]] + a[2]
>
> it's REALLY unnecessary to invoke the AG inequality to prove this is
> non-negative. Just complete the square:
>
> Factor@Expand[4*(g[2]^2 - f[2]^2)]
>
> (Sqrt[a[1]] - Sqrt[a[2]])^2
>
> Bobby
>
> -----Original Message-----
> From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
To: mathgroup at smc.vnet.net
> To: mathgroup at smc.vnet.net
> Sent: Wed, 14 May 2003 08:07:49 -0400 (EDT)
> Subject: [mg41348] [mg41299] Re: [mg41285] which one is greater than or equal?
>
> This looks very suspiciously like a homework question and also don't
> forget that this is a Mathematica (the computer program) and not an
> elementary mathematics list. On the other hand it is a nice problem
so
> I have a dilemma...
> O.K., I know what I shall do.
>
> 1. I shall try to use Mathematica (a little anyway)
> 2. I'll give an incomplete proof (although it will be pretty trivial
to
> finish it off).
>
> O.K. here we go. Of course the basic idea is to use the famous
> inequality involving the arithmetic mean and the geometric mean (the
> AG inequality). So I shall assume that you know it. Actually we only
> need to use the inequality for two variables, that is
>
> (a+b)/2>=Sqrt[a b]
>
> (where a>=0 and b >=0).
>
> In fact, it will be more convenient to use it in the form
>
> (a+b)>= 2 Sqrt[a b]
>
> Now we begin. First let us define two functions defining the left and
> right hand sides of your inequality:
>
> In[1]:=
> f[n_]:=Sum[Sqrt[a[i]],{i,1,n}]/n
>
> In[2]:=
> g[n_]:=Sqrt[Sum[a[i],{i,1,n}]/n]
>
> I claim that g[n] >= f[n] for every positive integer n.
>
> Let's first try if Mathematica can do this by itself. It obviously
> won't be able to work witha general n, so let's try a few cases:
>
> In[3]:=
> Timing[Simplify[g[2] >= f[2], {a[1] >= 0, a[2] >= 0}]]
>
> Out[3]=
> {0.21000000000000002*Second, True}
>
> In[4]:=
> Timing[Simplify[g[3] >= f[3], {a[1] >= 0, a[2] >= 0,
> a[3] >= 0}]]
>
> Out[4]=
> {0.5199999999999999*Second, True}
>
> In[5]:=
> Timing[Simplify[g[4] >= f[4], {a[1] >= 0, a[2] >= 0,
> a[3] >= 0, a[4] >= 0}]]
>
> Out[5]=
> {0.76*Second, True}
>
> Very good but it doesn't tell us how it did it. So let's try to do it
> "manually".
> Since we are of course assuming that everything in sight is
> non-negative all we need to show is that
> g[n]^2 >= f[n]^2. Well lets look at a few cases for small n:
>
> In[6]:=
> Expand[4*(g[2]^2 - f[2]^2)]
>
> Out[6]=
> a[1] - 2*Sqrt[a[1]]*Sqrt[a[2]] + a[2]
>
> well, this is obviously just
>
> a[1]+a[2]-2*Sqrt[a[1]]*Sqrt[a[2]]
>
> so by the AG inequality this is >= 0. Case n=2 done.
>
> Let's look at n=3:
>
> In[7]:=
> Expand[9*(g[3]^2 - f[3]^2)]
>
> Out[7]=
> 2*a[1] - 2*Sqrt[a[1]]*Sqrt[a[2]] + 2*a[2] -
> 2*Sqrt[a[1]]*Sqrt[a[3]] - 2*Sqrt[a[2]]*Sqrt[a[3]] + 2*a[3]
>
> This we can re-write in the form:
>
> (a[1] + a[2] - 2*Sqrt[a[1]]*Sqrt[a[2]]) + (a[1] + a[3] -
> 2*Sqrt[a[1]]*Sqrt[a[3]]) + (a[3] + a[2] - 2*Sqrt[a[3]]*Sqrt[a[2]])
>
> and of course by the AG inequality each summand is >=0 so we are done
> in this case.
>
> Well, I think this is ought to be more than enough.
>
> Andrzej Kozlowski
> Yokohama, Japan
> http://www.mimuw.edu.pl/~akoz/
> http://platon.c.u-tokyo.ac.jp/andrzej/
>
>
> On Tuesday, May 13, 2003, at 05:18 pm, Son Bui wrote:
>
>> can any one tell me which one is greater than? Please give your
proof.
>>
>> sqrt((a1+a2+...+an)/n) ? (sqrt(a1)+sqrt(a2)+...+sqrt(an))/n
>>
>> Bui
>>
>>
>>
>>
>
>
Andrzej Kozlowski
Yokohama, Japan
http://www.mimuw.edu.pl/~akoz/
http://platon.c.u-tokyo.ac.jp/andrzej/