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.
This may not be the best way to do it, but it's my favorite method.
Since we are trying to prove that:
$\displaystyle |a + b| \leq |a| + |b|$
It suffices to show that:
$\displaystyle |a+b|^2 \leq (|a| + |b|)^2$
$\displaystyle |a+b|^2 = (a+b)(a+b)$
$\displaystyle = a^2 + 2ab + b^2$
$\displaystyle = |a|^2 + 2ab + |b|^2$
$\displaystyle \leq |a|^2 + 2|a||b| + |b|^2$
$\displaystyle \leq (|a| + |b|)^2$
The one that everyone else might use is:
$\displaystyle |a+b| \leq |a| + |b|$
Notice that $\displaystyle -|a| \leq a \leq |a|$ and that $\displaystyle -|b| \leq b \leq |b|$.
When we add those two statements together, we get:
$\displaystyle -(|a| + |b|) \leq a+b \leq (|a| + |b|)$
by the absolute value property, you get:
$\displaystyle |a+b| \leq |a| + |b|$
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 $\displaystyle V$, the triangle inequality is:
$\displaystyle ||x+y|| \leq ||x|| + ||y||, \ \forall \ x,y \in V$
Since the real line is a normed vector space, then the triangle inequality for real numbers x and y is:
$\displaystyle |x+y| \leq |x| + |y|$
So... What triangle inequality should I be proving?