Re: Re: which one is greater than or equal?
- To: mathgroup at smc.vnet.net
- Subject: [mg41337] Re: [mg41299] Re: [mg41285] which one is greater than or equal?
- From: Bobby Treat <drmajorbob at mailblocks.com>
- Date: Thu, 15 May 2003 04:02:11 -0400 (EDT)
- Sender: owner-wri-mathgroup at wolfram.com
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
Subject: [mg41337] [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
>
>
>
>