B Is 2ab Always Less Than or Equal to a² + b² for All Integers a and b?

[PhysicsBoyMan]

a and b are integers

Prove that:
2ab <= a² + b²

I have tested various values for a and b and determined that the statement seems to be generally true. I'm having a hard time though constructing a formal proof.

It will not do to suppose the statement is wrong and then provide a counterexample. This would only prove that the statement is true for those particular values of a and b, and not for all values.

I learned about mathematical induction but I think that method is used to prove statements about a well-ordered set. a and b can be any integer so I don't think that would work.

I wondered what kind of statement may have simplified to produce a² + b² so I played with (a+b) * (a+b) which produced a² + b² + 2ab and also with (a-b) * (a-b) which produced a² + b² - 2ab. This looks similar to the original statement, and I feel like I may be onto some clues.

Without giving me the answer, I wonder if someone could give me a push in the right direction?

 

[fresh_42]

The last formula gives you the answer. What do you know about squares?

 

[PhysicsBoyMan]

The last formula gives you the answer. What do you know about squares?

Well I know that squares are always positive.

 

[fresh_42]

Then gather all together. Start with "all square are positive". Then substitute a certain square of one of your formulas and rearrange the terms..

 

[PhysicsBoyMan]

Then gather all together. Start with "all square are positive". Then substitute a certain square of one of your formulas and rearrange the terms..

I don't know what some of your comment means. I'm trying hard to figure it out. I'm looking at the term (a-b)² and I think that may be the "certain square" you are talking about. It's equal to a² + b² - 2ab though so I don't know what I could substitute it for. Maybe I'm on the wrong track here altogether.

 

[fresh_42]

I don't know what some of your comment means. I'm trying hard to figure it out. I'm looking at the term (a-b)² and I think that may be the "certain square" you are talking about. It's equal to a² + b² - 2ab though so I don't know what I could substitute it for. Maybe I'm on the wrong track here altogether.

No, you are done. The square is not negative ($≥ 0$) and all which remains, is to add $2ab$ on both sides.

 

[PhysicsBoyMan]

No, you are done. The square is not negative ($≥ 0$) and all which remains, is to add $2ab$ on both sides.

How is that a proof that 2ab is always less than or equal to a² + b² when a and b are integers?

 

[fresh_42]

How is that a proof that 2ab is always less than or equal to a² + b² when a and b are integers?

The only way to help you further is to type in the proof.

Because you can only derive true statements from true statements, it is a proof:
$0 ≤ (a - b)^2$ for all integers, and even for real numbers, too, is a true statement. The next step is to multiply the square which is of course also true. Shifting numbers by addition doesn't change the order, so you are allowed to add $2ab$ on both sides and get again a true statement, which is the one you want to have.

 

[PhysicsBoyMan]

Thanks. I didn't think about it like that. That is, starting with a separate true statement and transforming it into the one I want.

 

[PeroK]

Thanks. I didn't think about it like that. That is, starting with a separate true statement and transforming it into the one I want.

That's probably the simplest proof but you could have done it simply by contradiction as well:

Suppose $a^2 + b^2 < 2ab$ then

$a^2+b^2 -2ab <0$
And
$(a-b)^2 <0$

Which is a contradiction.

 

[PhysicsBoyMan]

d

That's probably the simplest proof but you could have done it simply by contradiction as well:

Suppose $a^2 + b^2 < 2ab$ then

$a^2+b^2 -2ab <0$
And
$(a-b)^2 <0$

Which is a contradiction.

Nice one Perok. Contradictions are a lot more intuitive often times.