Can Absolute Value Property be Proven for Real Numbers x and y?

[knowLittle]

Homework Statement

Prove that for every two real numbers x and y
$|x+y| \leq |x| + |y|$

Homework Equations

The Attempt at a Solution

There are three cases. The easiest ones is when they are both positive and negative.
The third one I have problems with.
The numbers have different sign. Say x>0 and y<0
Divide this into two subcases:
case 3.1
$x+y \geq 0$$|x| +|y| = x+(-y) = x-y$
Now, so far so good, but my book states the following.
$|x| +|y| = x+(-y) = x-y > x+y = |x+y|$
How is it possible that x-y be ever greater than x+y?

case 3.2
$x+y < 0$
This one is easy too.

 

[pasmith]

How is it possible that x-y be ever greater than x+y?

If $-y > y$, so that $0 > 2y$, ie. $y < 0$.

 

[knowLittle]

$-y > y$ would be false even if $y<0$. They would be equal not greater than each other.

I don't see the connection to the proof. Thanks for trying.

 

[pasmith]

$-y > y$ would be false even if $y<0$. They would be equal not greater than each other.

Let $y = -1$. Do you agree that $-y = -(-1) = 1 > -1 = y$?

 

[Ray Vickson]

Homework Statement

Prove that for every two real numbers x and y
$|x+y| \leq |x| + |y|$

Homework Equations

The Attempt at a Solution

There are three cases. The easiest ones is when they are both positive and negative.
The third one I have problems with.
The numbers have different sign. Say x>0 and y<0
Divide this into two subcases:
case 3.1
$x+y \geq 0$


$|x| +|y| = x+(-y) = x-y$
Now, so far so good, but my book states the following.
$|x| +|y| = x+(-y) = x-y > x+y = |x+y|$
How is it possible that x-y be ever greater than x+y?

case 3.2
$x+y < 0$
This one is easy too.

"How is it possible that x-y be ever greater than x+y?" Try x = 1 and y = -1. What is x-y? What is x+y?

 

[knowLittle]

Thank you, I was confused.