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- bobbym
- 2011-09-13 07:22:34
Hi;
I can use it a lot easier than explain it. Here is the site that taught me.
http://2000clicks.com/MathHelp/IneqMuir … ality.aspx
- anonimnystefy
- 2011-09-13 06:47:58
what is muirheads' inequality?
- bobbym
- 2011-09-13 03:41:43
Hi all;
- bobbym
- 2011-09-12 06:46:21
Hi;
why not?
Try a = 2, b = 2, c = 1.
I think this is where your error is.
You are okay after A)
Mistake made in B.
It was a pretty good idea though.
- anonimnystefy
- 2011-09-12 06:43:28
why not?
- bobbym
- 2011-09-12 06:42:03
Hi;
This statement is not true under the conditions of the problem.
- anonimnystefy
- 2011-09-12 06:31:20
hi guys
is this correct for (3):
- bobbym
- 2011-09-12 06:06:04
It is okay. I am not sure about the problem either.
- bob bundy
- 2011-09-12 05:15:52
Ok, my mistake.
Bob
- bobbym
- 2011-09-12 04:38:06
Hi Bob;
Positive numbers a,b, and c satisfy the inequality a+b+c>= abc. Prove that a^2+b^2+c^2>=(abc)^2
because that is do-able.
Not if you have
a = 2, b = 5 / 4, c = 2
a = 3, b = 1, c = 2
I am fairly sure the original problem #3 is true.
- bob bundy
- 2011-09-12 04:29:28
This is for grade 9 and the other two were fairly straight forward. Could the question be:
Positive numbers a,b, and c satisfy the inequality a+b+c>= abc. Prove that a^2+b^2+c^2>=(abc)^2
because that is do-able.
Bob
- bobbym
- 2011-09-11 22:13:11
Hi;
A little help.
3) Positive numbers a,b, and c satisfy the inequality a+b+c>= abc. Prove that a^2+b^2+c^2>=abc
By Muirheads inequality:
Therefore it is not mandatory that this can be done by the AMGM.
- bob bundy
- 2011-09-10 18:22:38
hi again,
(2) Like this:
and
So put these together and you get what you wanted.
Now for (3). edit: Many hours later. I cannot do this one yet. I've put out a general request for more brains.
Bob
- bob bundy
- 2011-09-10 18:09:06
hi fromwoodhouse,
Here's my proof for (1)
Consider the case m ≥ 0
In that case treat |m| as just m
but we know from the case that m and 1 + m > 0 so
Now consider the alternative case that m < 0
That means we can replace |m| with -m
this time we know 1 - m > 0 as m < 0 so
If a product is negative then one factor must be + and one must be - so
either m > 0 and (m + 1) < 0 but this contradicts the assumption that m < 0
or m < 0 and (m + 1) > 0 which leads to -1 < m < 0
Now to look at number (2). See next post.
Hope that helps
Bob
- reconsideryouranswer
- 2011-09-10 02:22:03
fromwodehouse wrote:
1) Prove: if n<0 and |m|= m-n/m+n , -1<m<0 or m>1
2) Given m = 4-x/3, n = x+3/4, p = 2-3x/5 , and m>n>p
Prove that the possible value of x is {-7/17<x<1}
fromwodehouse,
because of the Order of Operations,
you must use grouping symbols for these: