Simple Equation Does my proof work? Its the inequality equation $|a+b| \leq |a|+|b
| $ 
I managed this by cases.
Let $c = a$ and $d=b$ if $a>b $
let $c = b$ and $d = a$ if $b>a $
if $a=b$ let $a=c$ 
Hence we have $|c+d| \leq |c|+|d| $ where $c>d$ 
Case 1
if  $d > 0$   then $|c+d| = |c|+|d|$ $\forall c $
Case 2 
if $c < 0$ then $|c+d| = |c|+|d|$ $\forall d $
Case 3 if $ c \vee d =0$ then $|c+d| = |c|+|d|$
Case 4   $c > 0$  and $d < 0$ then $|c+d| < |c|+|d|$ $ \forall$ $c$ and $d$
But i want to solve this directly not by cases.
I think that it can be done by treating a and b as vectors described by two components up down ($y$) and left right ($x$) where $|a|$ is the length of the vector $\vec a$ described as:
$$|\vec a|=  \sqrt {x^2_a + y^2_a}$$
$$|\vec b|=  \sqrt {x^2_b + y^2_b}$$
And $|\vec a+\vec b|$ would be the length of $\vec a$ plus $\vec b$ or $|\vec a+\vec b|= \sqrt {(x_a +x_b)^2 + (y_a +y_b )^2 }$
thus $|\vec a+\vec b| \leq |\vec a|+|\vec b| $ = $ \sqrt {(x_a +x_b)^2 + (y_a +y_b )^2 } \leq \sqrt {x^2_a + y^2_a} + \sqrt {x^2_b + y^2_b}$ now since we cant have negative length we can square both sides.
$ (x_a +x_b)^2 + (y_a +y_b )^2  \leq  x^2_a + y^2_a + x^2_b + y^2_b + 2(x^2_b + y^2_b)^{1/2} (x^2_a + y^2_a)^{1/2}$
$ ( x_a x_b + y_a y_b) \leq (x^2_b + y^2_b)^{1/2} (x^2_a + y^2_a)^{1/2}$
rhs = $(x^2_a x^2_b  + y^2_a x^2_b + x^2_a y^2_b  + y^2_a y^2_b)$ ?
anyway from here it feels like im screwed on the plus side the right hand side is always $0 \geq  $ and the left side can be positive or negative. oddly whenever the x and y components of the vectors have a magnitude greater then or equal to 1 this seems to stand up fairly obviously its just when it has pieces under magnitude 1 that things get really ugly though only intuitively it feels obvious that when none of the components or either vector are $0$ the right side should be bigger then the left.
any ideas on a different way to approach this? thanks.
EDIT: i Originally did exactly what Rolighed has written down and came to the conclusion that $|x| |y| \geq |xy|$ as he did at the end of his proof but i tossed out the idea on the thought that i couldn't square both sides on the thought that $4 \geq -5$ but $4^2 < -5^2$
EDIT2: it turns out that this inequality can be squared because both sides are positive no matter what which makes if $ x \geq y$ and $ y \geq 0$  then $x^2 \geq y^2$  since we know both sides of our inequality must be 0 or bigger we can square them and do the easy proof as shown by Rolighed. My confusion of your explanation Rolighed how you come you could square both sides took me too long to realize why =(
 A: Well.. I learned this proof for the inequality.
Let $x,y\in\mathbb{R}^k$. Observe that 
$|x+y|^2=(x+y)(x+y)=|x|^2+|y|^2+2xy\le |x|^2+|y|^2+2|xy| \le|x|^2+|y|^2+2|x||y|=(|x|+|y|)^2$
In the last inequality, we have used the Cauchy/Schwarz Inequality ($|xy|\le|x||y|$). For $k>1$ multiplication is exchanged by vector product.
A: Here's one way to simplify your argument. I think it's inevitable to have to resort to some sort of case analysis.
Case 1: $b=0$. Obvious. (Right?)
Case 2: $b\neq0$. It's enough to prove that $|1+a/b|\leq1+|a/b|$, since if this inequality is true we can multiply by $|b|$ to get the inequality we want. Let $x=a/b$, just to save the trouble of writing $a/b$. So we want to prove that $|1+x|\leq1+|x|$.
Case 2(a): $x\geq0$. Easy -- $|x|=x$ and we get $1 + x\leq 1 + x$, which is true.
Case 2(b): $x\leq0$. In this case $|x|=-x$. We get $1 + x\leq 1 - x$, which, if we solve for $x$, is the same as saying that $2x \leq 0$, i.e., $x\leq0$.
Do you have an intuition for why this inequality should be true, and what it means?
