# Proving a vector inequality in R^n

1. Jan 28, 2009

### king vitamin

1. The problem statement, all variables and given/known data
In R^n show that:

$$\Vert\overrightarrow{x - y}\Vert \Vert\overrightarrow{x+y}\Vert \leq \Vert\overrightarrow{x}\Vert^{2} + \Vert\overrightarrow{y}\Vert^{2}$$

2. Relevant equations

My main attempts have either been from the Triangle Inequality:
$$\Vert\overrightarrow{x+y}\Vert \leq \Vert\overrightarrow{x}\Vert + \Vert\overrightarrow{y}\Vert$$

Or from attempts to implement the idea:
sqrt(a^2 + b^2) <= |a| + |b|

3. The attempt at a solution
Everytime I attempt to do this algebraically, the product on the left becomes a distributive catastrophe and I can't get anything sensible out of it, and attempting to square both sides to get rid of the square root just results in the same catastrophe on the RHS. I tried to represent the left hand side as a dot product but that involved cos(theta) which just complicates the problem. If anyone could point me in the right direction it would be appreciated.

2. Jan 28, 2009

### CompuChip

The triangle equality is all you need. Just square both sides and you're almost there already.

3. Jan 28, 2009

### king vitamin

I feel more stuck than before! Squaring the Triangle inequality obtains:

$$\Vert\overrightarrow{x+y}\Vert^{2} \leq \Vert\overrightarrow{x}\Vert^{2} + \Vert\overrightarrow{y}\Vert^{2} + 2\Vert\overrightarrow{x}\Vert\Vert\overrightarrow{y}\Vert$$

So do I now need to show that:

$$\Vert\overrightarrow{x - y}\Vert \Vert\overrightarrow{x+y}\Vert \leq \Vert\overrightarrow{x+y}\Vert^{2} - 2\Vert\overrightarrow{x}\Vert\Vert\overrightarrow{y}\Vert$$
? EDIT: I found a counterexample to this equation, so mark it off the list

Even more frustrating is this easy result from the Triangle Inequality:

$$\Vert\overrightarrow{x+y}\Vert\Vert\overrightarrow{x-y}\Vert \leq \Vert\overrightarrow{x}\Vert^{2} + \Vert\overrightarrow{y}\Vert^{2} + 2\Vert\overrightarrow{x}\Vert\Vert\overrightarrow{y}\Vert$$

I just can't get rid of the 2xy term!

Last edited: Jan 28, 2009
4. Jan 28, 2009

### CompuChip

Oohhh, now I see, there is a minus sign in one of them.
Then I think the triangle inequality doesn't help you much... sorry.

I did get it to work (more or less) through the dot product. You have to think about cos(theta) a bit, but eventually you will be able to use that a - b <= a if b >= 0 and sqrt(a^2 + b^2) <= |a| + |b| to get where you want.
I hope that's a better hint and you'll get there now.

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