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# MATHS Help please! Proof... watch

1. Prove that x^2 + y^2 + z^2 >/= xy + yz + zx

I've never understood the "prove that..." questions because how exactly do I prove that it is always correct no matter what?!

I started by multiplying both sides by 2, and trying use the idea that (x+y+z)^2 = x^2 + y^2 + z^2 + 2(xy + yz + zx) in conjunction with that to try to do something, but I only end up with alternative inequalities, no proof.
2. Set up a statement in which x, y, z∈ℝ

Thus the following is true
(x - y)2 + (x - z)2 + (y - z)2 ≥ 0
Square numbers lead only to positive real numbers

Expand and simplify to get
2x2 + 2y2 + 2z2 - 2xy - 2xz - 2yz ≥ 0

Divide by 2 and rearrange.
x2 + y2 + z2 ≥ xy + yz + zx
3. (Original post by RMNDK)
Set up a statement in which x, y, z∈ℝ

Thus the following is true
(x - y)2 + (x - z)2 + (y - z)2 ≥ 0
Square numbers lead only to positive real numbers

Expand and simplify to get
2x2 + 2y2 + 2z2 - 2xy - 2xz - 2yz ≥ 0

Divide by 2 and rearrange.
x2 + y2 + z2 ≥ xy + yz + zx
Oh my word I feel stupid... Thank you so much! I don't understand why I didn't see this though.

Wait so you're allowed to end with the inequality, I thought you had to start with it?!
4. (Original post by ComputerMaths97)
Oh my word I feel stupid... Thank you so much! I don't understand why I didn't see this though.

Wait so you're allowed to end with the inequality, I thought you had to start with it?!

Don't be, I wouldn't have figured it out easily until my teacher showed it me.

You start and end with the inequality. Yeah I did start with the inequality? You start and end with it in this proof.
5. (Original post by RMNDK)
Set up a statement in which x, y, z∈ℝ

Thus the following is true
(x - y)2 + (x - z)2 + (y - z)2 ≥ 0
Square numbers lead only to positive real numbers

Expand and simplify to get
2x2 + 2y2 + 2z2 - 2xy - 2xz - 2yz ≥ 0

Divide by 2 and rearrange.
x2 + y2 + z2 ≥ xy + yz + zx
Please don;t post full solutions - one of the forum rules

It does not help the OP
6. (Original post by ComputerMaths97)
Prove that x^2 + y^2 + z^2 >/= xy + yz + zx

I've never understood the "prove that..." questions because how exactly do I prove that it is always correct no matter what?!

I started by multiplying both sides by 2, and trying use the idea that (x+y+z)^2 = x^2 + y^2 + z^2 + 2(xy + yz + zx) in conjunction with that to try to do something, but I only end up with alternative inequalities, no proof.
Where did you get this question from? I need more like this
7. (Original post by RMNDK)
Don't be, I wouldn't have figured it out easily until my teacher showed it me.

You start and end with the inequality. Yeah I did start with the inequality? You start and end with it in this proof.
No but you just instantly said:

"(x-y)^2 + (x-z)^2 + (y-z)^2 >/= 0"

Yet I have no idea how you just appeared there. Like I know that it's true, but I never thought to just randomly begin with another fact about integers lol xD
8. I know how you feel. It's a bit weird how one just states a proposition out of almost thin air. It's called a lemma, a theorem on the side if you will. You state something separate to the proof and once you can prove the lemma, it gets treated as if it's a universal proposition.

If you want further reading, look at how to prove the Addition Rule for Differentiation. Another one of those 'instantly said' proofs.
9. (Original post by ComputerMaths97)
No but you just instantly said:

"(x-y)^2 + (x-z)^2 + (y-z)^2 >/= 0"

Yet I have no idea how you just appeared there. Like I know that it's true, but I never thought to just randomly begin with another fact about integers lol xD
When you do enough of these things, they become trivially obvious. Including the original proposition.

Oh, and the quick way I use for these things is AM-GM.
10. (Original post by Renzhi10122)
When you do enough of these things, they become trivially obvious. Including the original proposition.

Oh, and the quick way I use for these things is AM-GM.
Ah okay so just one of those things that come with practise?

AM-GM?
11. (Original post by ComputerMaths97)
Ah okay so just one of those things that come with practise?

AM-GM?
Pretty much. Arithmetic mean geometric mean inequality, google it.
12. (Original post by Renzhi10122)
Pretty much. Arithmetic mean geometric mean inequality, google it.
Oh so this is a specific group of mathematical facts essentially, that are commonly known so used to answer questions like this?

Could you state that you know the AM-GM rule is always true, if at some point through your proof you reach an example of an AM-GM inequality and finish there? Or would you have to specifically prove it each time?

Very helpful though, thanks! I feel, if I knew this when I started the question I would've found it easier now I've read a few pages on it. Is there somewhere I can find out about more mathematical facts that are often used in proofs? Or is this just a pattern I was meant to eventually realise?
13. (Original post by ComputerMaths97)
Oh so this is a specific group of mathematical facts essentially, that are commonly known so used to answer questions like this?

Could you state that you know the AM-GM rule is always true, if at some point through your proof you reach an example of an AM-GM inequality and finish there? Or would you have to specifically prove it each time?

Very helpful though, thanks! I feel, if I knew this when I started the question I would've found it easier now I've read a few pages on it. Is there somewhere I can find out about more mathematical facts that are often used in proofs? Or is this just a pattern I was meant to eventually realise?
What do you mean by the bold?

It depends on the situation. Never in an A level maths exam would I recommend using an unorthodox method. In BMO and upwards, use it as you wish, and it's also fine in STEP. You would never need to prove it each time, it's more the question of using it or not.

What's this from? Learning more theorems takes time and effort, so it may be pointless to learn more if this is just a one off question and you don't really like olympiads.
14. (Original post by Renzhi10122)
What do you mean by the bold?

It depends on the situation. Never in an A level maths exam would I recommend using an unorthodox method. In BMO and upwards, use it as you wish, and it's also fine in STEP. You would never need to prove it each time, it's more the question of using it or not.

What's this from? Learning more theorems takes time and effort, so it may be pointless to learn more if this is just a one off question and you don't really like olympiads.
I do like olympiads and I will be sitting STEP so where can I learn more theorems that I can learn? I've got the time lol
15. (Original post by ComputerMaths97)
I do like olympiads and I will be sitting STEP so where can I learn more theorems that I can learn? I've got the time lol
Here's a good starting point then
http://www.ukmt.org.uk/docs/BMO%20Pr...on%20Sheet.pdf
16. (Original post by Renzhi10122)
... it's also fine in STEP.
In general its fine, but you do get the occasional question where using AM-GM is dangerously close to begging the question. One question explicitly asked you to prove it (using Jensen's inequality) - that's extreme, but I think if asked to show something like quoting AMGM would be pushing your luck.

(Original post by ComputerMaths97)
..
For what it's worth, with this question my thought process was:

"I really want to have 2xy+2yz+2zx on the RHS, so I can start getting some (A-B)^2 type terms. What happens if I multiply by 2? Then 2x^2+2y^2+2z^2 doesn't look good {pause to think}, but actually, something like (x-y)^2 contributes two x^2 type terms, so maybe it's OK. {pause}. Yes, (x-y)^2+(x-z)^2+(z-x)^2 is going to work".
17. (Original post by DFranklin)
In general its fine, but you do get the occasional question where using AM-GM is dangerously close to begging the question. One question explicitly asked you to prove it (using Jensen's inequality) - that's extreme, but I think if asked to show something like quoting AMGM would be pushing your luck.

For what it's worth, with this question my thought process was:

"I really want to have 2xy+2yz+2zx on the RHS, so I can start getting some (A-B)^2 type terms. What happens if I multiply by 2? Then 2x^2+2y^2+2z^2 doesn't look good {pause to think}, but actually, something like (x-y)^2 contributes two x^2 type terms, so maybe it's OK. {pause}. Yes, (x-y)^2+(x-z)^2+(z-x)^2 is going to work".
Ok, STEP may be a bit far. I was once told that you should try to aim to use theorems that the 'above average' person taking the exam knows (I once wanted to use an IMO question to solve a BMO one...), and AM-GM just about makes it, though of course if the whole point of the question is about proving it, then it's a very bad idea.
18. (Original post by Renzhi10122)
Ok, STEP may be a bit far. I was once told that you should try to aim to use theorems that the 'above average' person taking the exam knows (I once wanted to use an IMO question to solve a BMO one...), and AM-GM just about makes it, though of course if the whole point of the question is about proving it, then it's a very bad idea.
It's the last part of what you say that's the issue, which i don't think is specific to STEP at all (*). It's very much a question of judgement about what is going to end up "a very bad idea".

(*) The one "STEP specific" thing is that for most people STEP is the first exam where even the possibility of using off-syllabus results occurs (e.g. L'Hopital). So it's the first time they have to start thinking about "is it really OK for me to quote this result here?"

Of course, the BMO/IMO questions are kind of extreme the other way. AIUI, you can use any published result, even if it completely trivialises the question. Again AIUI, Mihăilescu's proof of Catalan's conjecture in 2002 "squashed" a number of old problems, since they then fell to 1 line solutions quoting the newly proved result.

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