Proving Inequality in Inner Product Spaces

[ChickysPusss]

Homework Statement

First I'd like to state the meaning of my notations
x = (x0,x1,x2...xn)
y = (y0,y1,y2...yn)
|x| = absolute value of x
||x|| = Normal of x
<x,y> = Inner Product of x and y

I have to prove the following

|<x1,y1> - <x2,y2>| ≤ ||x1 - x2||*||y1|| + ||x2||*||y1-y2||

Homework Equations

Applicable Axioms of Normals and Inner Products
||x|| = √(<x,x>)
<x + z,y> = <x,y> + <z,y>

The Attempt at a Solution

I tried expanding the right hand side as such:
||x1 - x2|| = √(<x1-x2,x1-x2>) = √(<x1,x1> + 2*<x1,-x2> + <-x2,-x2>)
||x2|| = √(<x2,x2>)

I did similarly for the y values, and I'm not seeing anything that pops out to me as a solution to this proof, nothing seems to cancel, and no axioms seem to make this work in a general sense. I guess what I need...is a HINT.

 

[LCKurtz]

Hint: Try subtracting and adding $\langle x_2,y_1\rangle$.

 

[ChickysPusss]

Thank you, but I still don't see how that's applicable, everything's inside an absolute value or a square root so I'm not sure how to get in there? Can I have a double hint?

 

[LCKurtz]

Don't you have the Cauchy-Schwartz inequality $|\langle x,y\rangle|\le \|x\|\|y\|$ for an inner product space? That plus the first hint should do it...

[Edit, added] And, of course, the triangle inequality

 

[ChickysPusss]

LCKurtz there's no easy way to put this, but you are THE MAN. Thank you so much, here's the solution I came to thanks to your hints. I really should've came to it easier, but what can you do.

|<x1,y1> - <x2,y2>| ≤ ||x1 - x2||*||y1|| + ||x2||*||y1-y2||

Using the first hint so graciously given me by LCKurtz, I added and subtracted <x2,y1> on the left hand side.

This gives:
|<x1,y1> - <x2,y1> - <x2,y2> + <x2,y1>|
Then by an axiom listed above, we can bring this to:
|<x1 -x2,y1> + <y1-y2,x2>|

Then working on the right hand side, and using the memory jump-starter so humbly given by Zeus Incarnate, Mr. LCKurtz (Talking about the Cauchy-Schwartz inequality).
||x1-x2||*||y1|| >= |<x1-x2,y1>|
||y1-y2||*||x2|| >= |<y1-y2,x2>|

This means the right hand side as it is is greater than or equal to |<x1-x2,y1> + <y1-y2,x2>|, also known as... DUN DUN DUNNNNN the left hand side! So clearly, this is the end of the proof. A shout out to my boy LCKurtz for his help, thanks.