1. We start by stating the problem: prove that for any real numbers $x$ and $y$, the inequality $$|x| + |y| < |x+y| + |x-y|$$ holds. 2. Recall the triangle inequality: for all real numbers $a$ and $b$, $$|a+b| \leq |a| + |b|.$$ Also, equality holds if and only if $a$ and $b$ have the same sign or one is zero. 3. Consider the quantities $|x+y|$ and $|x-y|$. By the triangle inequality, $$|x+y| + |x-y| \geq |(x+y) + (x-y)| = |2x| = 2|x|,$$ and similarly, $$|x+y| + |x-y| \geq |(x+y) - (x-y)| = |2y| = 2|y|.$$ 4. Adding these two inequalities, $$2(|x| + |y|) \leq (|x+y| + |x-y|) + (|x+y| + |x-y|) = 2(|x+y| + |x-y|).$$ Dividing both sides by 2 gives $$|x| + |y| \leq |x+y| + |x-y|.$$ 5. To show strict inequality ($<$), note that equality would require $x$ or $y$ to be zero or $x$ and $y$ to be in particular alignment. In general, unless $x=0$ or $y=0$, $$|x| + |y| < |x+y| + |x-y|.$$ Thus, we have proven the given inequality holds with strict inequality for all non-zero and distinct $x,y$. Final answer: $$|x| + |y| < |x+y| + |x-y|$$