# Proof of triangle inequality.

• September 8th 2008, 02:37 PM
reagan3nc
Proof of triangle inequality.
Prove:
The sum of the lengths of any two sides of a triangle is greater than the length of the third side. This is known as the triangle inequality.

I need some help with this proof.

Thanks in advance for any help.
• September 9th 2008, 05:51 AM
Aryth
This may not be the best way to do it, but it's my favorite method.

Since we are trying to prove that:

$|a + b| \leq |a| + |b|$

It suffices to show that:

$|a+b|^2 \leq (|a| + |b|)^2$

$|a+b|^2 = (a+b)(a+b)$

$= a^2 + 2ab + b^2$

$= |a|^2 + 2ab + |b|^2$

$\leq |a|^2 + 2|a||b| + |b|^2$

$\leq (|a| + |b|)^2$

The one that everyone else might use is:

$|a+b| \leq |a| + |b|$

Notice that $-|a| \leq a \leq |a|$ and that $-|b| \leq b \leq |b|$.

When we add those two statements together, we get:

$-(|a| + |b|) \leq a+b \leq (|a| + |b|)$

by the absolute value property, you get:

$|a+b| \leq |a| + |b|$
• September 9th 2008, 08:01 AM
Krizalid
Aryth, you missunderstood the question, read it again. The original one doesn't involve the absolute value.
• September 9th 2008, 01:44 PM
Aryth
Ah, well, the triangle inequality I was using was:

The triangle inequality states that for any triangle, the length of a given side must be less than or equal to the sum of the other two sides but greater than or equal to the difference between the two sides.

In a normed vector space $V$, the triangle inequality is:

$||x+y|| \leq ||x|| + ||y||, \ \forall \ x,y \in V$

Since the real line is a normed vector space, then the triangle inequality for real numbers x and y is:

$|x+y| \leq |x| + |y|$

So... What triangle inequality should I be proving?