abs(x-y)>abs(x)-abs(y)
The > should be ≧ because they are often equal, e.g., |3-2| = |3|-|2|,
so I will change your problem to
Prove that |x-y| ≧ |x|-|y|
xy ≦ |xy| because they are equal if either x or y is 0 or if
they are both positive or both negative, and if
they have opposite signs the left side is
negative and the right side is positive, so the
inequality holds in all cases.
-2xy ≧ -2|xy| We multiplied both sides by -2, which reversed
the inequality.
x²-2xy+y² ≧ x²-2|xy|+y² We added x²+y² to both sides.
x²-2xy+y² ≧ |x|²-2|x||y|+|y|² On the right, we replaced x² by |x|², y²
by |y|² and |xy| by |x||y|
(x-y)² ≧ (|x|-|y|)² We factored both sides.
|x-y| ≧ |x|-|y| We took non-negative square roots of both sides.
Edwin
