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  Discussion about math, puzzles, games and fun.   Useful symbols: √ ∞ ≠ ≤ ≥ ≈ ⇒ ∈ Δ θ ∴ ∑ ∫ π -

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Topic review (newest first)

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.

smile

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:




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