The absolute value of a sum of two numbers is less than or equal to the sum of the absolute values of two numbers I am trying to prove (or this could be false) that 
$|x+y| \leq |x| + |y|$
 A: Note that $|a| \geq a$, $|b| \geq b$. Now, 
$a^2+b^2+2|a||b| \geq a^2+b^2 + 2ab \implies (|a|+|b|)^2 \geq (a+b)^2 \implies |a|+|b| \geq |a+b|$.
A: If $x=0$ or $y=0$, it's obvious that $|x + y| = |x| + |y|$.
Let's assume that both $x$ and $y$ are not zero, and $|x| \le |y|$, then $|x/y| \le 1$, and
$$|x + y|/|y| = |1 + x/y| = 1 + x/y \le 1 + |x/y| = (|x| + |y|)/|y|$$
Therefore $|x + y| \le |x| + |y|$ 
A: If $x\ge 0; y\ge $ then $x = |x|; y=|y|$ and $x + y \ge 0$ so $|x+y| = x+y $ are the result is obvious.  
If $x < 0$ and $y<0$ then $x=-|x|$ and $y=-|y|$ and $x+y  <0 $ so $|x+y| =-x-y = |x| + |y|$
If $x \ge 0$ and $y  < 0$ then $x+y = x -|y| < x = |x| < |x| +|y|$
If $x <0$ and $y \ge 0$ then $x + y = y -|x| < y = |y| < |x|+|y|$.
Or to put it more simply... if x and y are the same sign you add the absolute values.  If the signs are different then you subtract the absolute values and set positive.  A difference of positive values is always smaller than a sum, so $|x + y| = ||x| \pm |y||\le |x| + |y|$
