Sum of absolute values and the absolute value of the sum of these values? I'm working on a proof and I need some help with this:
I determined that for some situations ($x$ or $y$ are negative but not both):
$|x| + |y| > x + y$
How can I conclude using that statement the fact that:
$|x| + |y| > |x + y|$
 A: You can try considering $|a+b|^2$, then using the property that $x^2 \geq 0$ for all x, you can obtain the desired inequality. Also your conclusion should be $|x|+|y| \geq |x+y|$. Notice that the inequality is not strict. (For example, if $x=y=0$ then certainly 
$|0|+|0| > |0+0|$ is false.) Another way is to use the fact that $-|a| \leq a \leq |a|$ for all numbers a (doing this for both $a$ and $b$, and then manipulating the inequalities, we can achieve what you want), however, this second route assumes that you at least are a little bit familiar with the rules of inequalities, particularly the rules regarding inequalities with absolute values.
As an example of how we would apply the squaring technique, we can do the following:
$$ 0 \leq |a+b|^2  = (a+b)(a+b)  = a^2 +2ab +b^2.$$
Now since $a^2 \geq 0$ is always true we can say that
$$a^2 +2ab +b^2 = |a|^2 +2ab +|b|^2.$$ Now we want to try to "force" an inequality. That is, we will replace $2ab$ with  $2|a||b|$. If $a$ and $b$ are both greater that $0$ then nothing would change, however, if they are not both greater than $0$ we would get the following:
$$
|a|^2 +2ab +|b|^2 \leq |a|^2 +2|a||b| +|b|^2
$$ notice that we have broken the chain of equal signs and forced an inequality. (There are still some more steps to do for you. Hint: What is $(|a|+|b|)^2$?)
A: First method:
Since $$|x+y|^2=(x+y)^2=x^2+y^2+2xy \leq x^2+y^2+2|x| |y|=(|x|+|y|)^2$$ we have $$|x+y| \leq |x| + |y| $$ which is equality if and only if : $$xy=|x||y|=|xy|$$ if and only if $x$ and $y$  ar both negatives or both positives. This means taht if $x$ and $y$ hav'nt the same sign we have:
$$|x+y| < |x| + |y|$$
But $x+y \leq |x+y|$  gives also $x+y < |x|+|y|$
Second method: 
We have $$(|x|+|y|)-(x+y)=(|x|-x)+(|y|-y)$$
for all real number $a$ we have  $|a|-a \geq 0$, and equality holds if and only if $a$ is positive. Since $x$ or $y$ is'nt positive one of the numbers $|x|-x$ or $|y|-y$  is $ >0 $ so $|x|+|y| > x+y$
