Proof

fromwodehouse · 2011-09-08 22:35:04

we had a test and i got these three numbers wrong because i really dont understand a single thing about it and im still in grade 9. and since it bugging me i want to know how to do these things. Thanks ahead!

m-n
1) Prove: if n<0 and |m|= ------ , -1<m<0 or m>1
m+n

4-x x+3 2-3x
2) Given m = ------- , n = ------- , p = ------ , and m>n>p
3 4 5

7
Prove: the possible value of x is - <x <1
17

3) Positive numbers a,b, and c satisfy the inequality a+b+c> abc. Prove that a^2+b^2+c^2>abc

bobbym · 2011-09-08 22:59:58

Hi;

1 and 2 are totally unreadable on my browser. I am looking at 3.

fromwodehouse · 2011-09-08 23:08:59

okay thanks, 1 and 2 have fractions, i'll try to re write it again

fromwodehouse · 2011-09-08 23:15:27

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}

Bob · 2011-09-09 00:28:59

hi bobbym

I'm seeing

1) Prove: if n<0 and

-1<m<0 or m>1

and 2) Given

and m>n>p

Prove: the possible value of x is

Bob

Last edited by Bob (2011-09-09 00:33:28)

bobbym · 2011-09-09 00:39:13

Hi Bob;

That looks right!

Bob · 2011-09-09 00:46:44

So now we've done the hard bit we just need to do them.

Bob

reconsideryouranswer · 2011-09-09 04:22:03

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:

Bob · 2011-09-09 20:09:06

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

Last edited by Bob (2011-09-09 20:17:33)

Bob · 2011-09-09 20:22:38

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

Last edited by Bob (2011-09-10 03:38:52)

bobbym · 2011-09-11 00:13:11

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 · 2011-09-11 06:29:28

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 06:38:06

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 · 2011-09-11 07:15:52

Ok, my mistake.

Bob

bobbym · 2011-09-11 08:06:04

It is okay. I am not sure about the problem either.

anonimnystefy · 2011-09-11 08:31:20

hi guys

is this correct for (3):

bobbym · 2011-09-11 08:42:03

Hi;

This statement is not true under the conditions of the problem.

anonimnystefy · 2011-09-11 08:43:28

why not?

bobbym · 2011-09-11 08:46:21

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.

bobbym · 2011-09-12 05:41:43

Hi all;

anonimnystefy · 2011-09-12 08:47:58

what is muirheads' inequality?

bobbym · 2011-09-12 09:22:34

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

Math Is Fun Forum

#1 2011-09-08 22:35:04

Proof

#2 2011-09-08 22:59:58

Re: Proof

#3 2011-09-08 23:08:59

Re: Proof

#4 2011-09-08 23:15:27

Re: Proof

#5 2011-09-09 00:28:59

Re: Proof

#6 2011-09-09 00:39:13

Re: Proof

#7 2011-09-09 00:46:44

Re: Proof

#8 2011-09-09 04:22:03

Re: Proof

#9 2011-09-09 20:09:06

Re: Proof

#10 2011-09-09 20:22:38

Re: Proof

#11 2011-09-11 00:13:11

Re: Proof

#12 2011-09-11 06:29:28

Re: Proof

#13 2011-09-11 06:38:06

Re: Proof

#14 2011-09-11 07:15:52

Re: Proof

#15 2011-09-11 08:06:04

Re: Proof

#16 2011-09-11 08:31:20

Re: Proof

#17 2011-09-11 08:42:03

Re: Proof

#18 2011-09-11 08:43:28

Re: Proof

#19 2011-09-11 08:46:21

Re: Proof

#20 2011-09-12 05:41:43

Re: Proof

#21 2011-09-12 08:47:58

Re: Proof

#22 2011-09-12 09:22:34

Re: Proof

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