Show that this inequality is true for all x, y ε R

[tomcruisex]

Homework Statement

This is part of a question on absolute convergence on series. The following equation is given as a hint. It says that before answering the question on series I should prove that |xy| <= 1/2(|x|^2 + |y|^2) for any x,y ε R

Homework Equations

The Attempt at a Solution

I know that this is the same as |x||y| <= 1/2(|x||x| + |y||y|). That is about all I am able to do to manipulate this equation. I tried solving for x or y but found it is not possible to separate them out. So I can't see how to prove this other than -
letting x = 0, then showing the equation is true for a few values of y
letting x = 1, then showing the equation is true for a few values of y
etc...

And then saying the equation is true 'by induction'. Surely there is a better, less awkward, way to show that |xy| <= 1/2(|x|^2 + |y|^2) for any x,y ε R

 

[PeterO]

Homework Statement

This is part of a question on absolute convergence on series. The following equation is given as a hint. It says that before answering the question on series I should prove that |xy| <= 1/2(|x|^2 + |y|^2) for any x,y ε R

Homework Equations

The Attempt at a Solution

I know that this is the same as |x||y| <= 1/2(|x||x| + |y||y|). That is about all I am able to do to manipulate this equation. I tried solving for x or y but found it is not possible to separate them out. So I can't see how to prove this other than -
letting x = 0, then showing the equation is true for a few values of y
letting x = 1, then showing the equation is true for a few values of y
etc...

And then saying the equation is true 'by induction'. Surely there is a better, less awkward, way to show that |xy| <= 1/2(|x|^2 + |y|^2) for any x,y ε R

COllect all the terms on one side, and that side will be able to be expressed as a perfect square. I would always multiply each side by 2 to get rid of the nuisance fraction first.

 

[tomcruisex]

Ok, if I collect the terms on one side I get
|x|^2 -2|xy| + |y|^2 >= 0

which is
(|x| - |y|)^2 >= 0

So I can say
|x| - |y| >= 0

But this isn't true as you could have, for example, x = 5 and y = 7. Am I missing something here?

 

[PeterO]

Ok, if I collect the terms on one side I get
|x|^2 -2|xy| + |y|^2 >= 0

which is
(|x| - |y|)^2 >= 0

So I can say
|x| - |y| >= 0

But this isn't true as you could have, for example, x = 5 and y = 7. Am I missing something here?

You were going fine.

Look at your "which is" line

You have (...)² >= 0 I didn't fill in the bracket on purpose.

Tell me about the value of the square of anything involving real numbers.

 

[tomcruisex]

All I can think of is that the square of a real number will be a positive finite real number..?

So it doesn't matter what is in the brackets it will always be greater than or equal to zero and therefore (|x|-|y|)^2 >=0 or basically (...)^2 >= 0 is true. Is that correct?

 

[tomcruisex]

I am having trouble applying this 'hint' to the original question. The original question was -

Show that if $$\sum_{n=1}^{\infty}a_n^2$$ and $$\sum_{n=1}^{\infty}b_n^2$$ are convergent then $$\sum_{n=1}^{\infty}a_nb_n$$ is absolutely convergent.

Hint: Show |xy| <= 1/2(|x|^2 + |y|^2) for any x,y ε R

So I have shown that the hint is true but I can't see how I can just substitute in the series $$a_n$$ and $$b_n$$ for x and y as the |xy| cannot be represented by $$\sum_{n=1}^{\infty}a_nb_n$$ as that would be implying that
$$\sum_{n=1}^{\infty}a_n$$*$$\sum_{n=1}^{\infty}b_n$$ = $$\sum_{n=1}^{\infty}a_nb_n$$, which it isn't as multiplication of series doesn't work like that. Anyone know how to apply this hint to the original question?

 

[PeterO]

All I can think of is that the square of a real number will be a positive finite real number..?

So it doesn't matter what is in the brackets it will always be greater than or equal to zero and therefore (|x|-|y|)^2 >=0 or basically (...)^2 >= 0 is true. Is that correct?

That is the logic. The technique of collecting all terms on one side of an equation/inequation is a common approach.

 

[tomcruisex]

Cheers for the help mate. Any idea how to apply this hint to the original question in post #6?

 

[verty]

That hint is really all you need. Perhaps think about what relevance it has? (I'm being terse on purpose, you should know enough to answer this)