Prove that |x+y| <= |x| + |y|

Assume x and y are real numbers prove that the absolute(positive) value of x+y is less than or equal to the absolute value of x plus the absolute value of y.

Or:

Prove that |x+y| <= |x| + |y|

This proof has 4 parts:

PART 1
Assume x < 0 and y < 0 then x + y< 0, so that |x+y| = (x+y) or -(x+y) in the first case (x+y)=|x|+|y| and in the second case -(x+y)<|x|+|y| thus |x+y|<=|x|+|y|

Part 2
Assume x < 0 and y >= 0 then since x<0 then |x+y| < y < |x| + |y|

thus |x+y| < |x|+|y|

Part 3
Assume y < 0 and x >= 0 then since y<0 |x+y| < x < |x| + |y|

thus |x+y| < |x|+|y|

Part4
Assume x > 0 and y > 0 then |x+y| = x + y = |x| + |y|

thus |x+y|=|x|+|y|