The Student Room Group

Prove an inequality

Problem #12
image-b032d7ff-baef-4115-9d72-f5602f09d1f08035275979282684010-compressed.jpg.jpeg

According to the book it should be proven this way.
image-1f8df76e-f7a7-4f40-af00-20d2051d42375980074881344945456-compressed.jpg.jpeg

How do you know that you're supposed to do it that way?
I know how to get there when I know what they expect from me but if I don't have the solutions I do different operations.
image-79080874-fa56-479a-9977-4f194df42c7c4323092503784637277-compressed.jpg.jpeg

Any advice would be appreciated.
[quote(Original post by pgxrcix)]Problem #12
image-b032d7ff-baef-4115-9d72-f5602f09d1f08035275979282684010-compressed.jpg.jpeg

According to the book it should be proven this way.
image-1f8df76e-f7a7-4f40-af00-20d2051d42375980074881344945456-compressed.jpg.jpeg

How do you know that you're supposed to do it that way?
I know how to get there when I know what they expect from me but if I don't have the solutions I do different operations.
image-79080874-fa56-479a-9977-4f194df42c7c4323092503784637277-compressed.jpg.jpeg

Any advice would be appreciated.
One thing you could do is start with what you are trying to prove and get to a statement you know is true. The sequence of logic written on the paper is an example of this. What is useful is that you can reverse the logic and you then have a proof.
Reply 2
Original post by DarthYoda
One thing you could do is start with what you are trying to prove and get to a statement you know is true. The sequence of logic written on the paper is an example of this. What is useful is that you can reverse the logic and you then have a proof.


Does (a-b)^2 >= 0 prove the statement? Or are the square roots in the brackets necessary in order to prove the statement?
Original post by pgxrcix
Does (a-b)^2 >= 0 prove the statement? Or are the square roots in the brackets necessary in order to prove the statement?

You can start at (a+b)^²>=0 and then follow the steps backwards and it will be a valid proof.
Reply 4
The problem with your working is you assumed the proof to be true straight from the start
Reply 5
Original post by Sinnoh
The problem with your working is you assumed the proof to be true straight from the start


That's a method called direct proof. There's no contradiction, nor a possibility to list all possible cases.
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Reply 6
Original post by DarthYoda
You can start at (a+b)^²>=0 and then follow the steps backwards and it will be a valid proof.


Thank you! That answers my question.
Reply 7
Original post by pgxrcix
That's a method called direct proof. There's no contradiction, nor a possibility to list all possible cases.


So in this case, it's a given fact that (a+b)20(a + b)^2 \geq 0, and you're working back from that.

It says you use things established to be true, so you can't use the statement you're being asked to prove. In that example, they don't work backwards from the statement, they use known facts - like how integers before and after n are n + 1 and n - 1. That doesn't need proof, that's just a given.
This is the AMGM inequality and you can prove it by many ways: (a-b)^2 >= 0; (root a - root b)^2 >=0; or even induction, etc.

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