# Thread: Real Analysis: proving the triangle inequalities

1. ## Real Analysis: proving the triangle inequalities

I am trying to prove one of the triangle inequalities.

[] are going to be my absolute value symbols, sorry, i don't have mathematica on this computer with the symbols.

I am proving that [[x]-[y]]<=[x-y].

i tried this: we know -[x]<=x<=[x] and -[y]<=y<=[y]
if i subtract these i get: -[x]--[y]<=x-y<=[x]-[y]
then you simplify and get -([x]-[y])<=x-y<=([x]-[y])
so, [x-y]<=[[x]-[y]].
I thought i was on the right track but then i realized that the inequality was pointing in the wrong direction. Did i make some computational error or am i going about this the wrong way? I appreciate any hints on where I might start. thanks!

2. -|x|<=x<=|x|
-|y|<=-y<=|y|
|x-y|-|y|<=|x|
insert x=y-z and you get:
|z|-|y|<=|y-z|=|z-y|
Q.E.D

3. $\displaystyle |x| = |x - y + y| \le |x - y| + |y|$

$\displaystyle \implies |x - y| \ge |x| - |y|$ ...........(1)

Also, $\displaystyle |y| = |x - (x - y)| \le |x| + |x - y|$

$\displaystyle \implies |x - y| \ge |y| - |x|$ ............(2)

By (1) and (2) we see that

$\displaystyle -|x - y| \le |x| - |y| \le |x - y|$ and the conclusion follows (by the fact that $\displaystyle -b \le a \le b \implies |a| \le b$)