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

1. Oct 2, 2011

### tomcruisex

1. The problem statement, all variables and given/known data
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

2. Relevant equations

3. 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
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Oct 2, 2011

### PeterO

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.

3. Oct 2, 2011

### 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?

4. Oct 2, 2011

### PeterO

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.

5. Oct 2, 2011

### tomcruisex

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

So it doesnt 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?

Last edited: Oct 2, 2011
6. Oct 2, 2011

### tomcruisex

I am having trouble applying this 'hint' to the original question. The original question was -
So I have shown that the hint is true but I cant 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?

Last edited: Oct 2, 2011
7. Oct 2, 2011

### PeterO

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

8. Oct 2, 2011

### tomcruisex

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

9. Oct 3, 2011

### 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)