To prove that $||x| - |y|| \leq |x - y|$ without using the triangle inequality, we can proceed as follows:

Let's denote $x$ and $y$ as real numbers.

Case 1: $x \geq y \geq 0$ or $y \geq x \geq 0$

Assume $x \geq y \geq 0$. Then:

$|x| = x, \quad |y| = y, \quad \text{and} \quad |x - y| = x - y$

So, the left side of the inequality becomes:

$||x| - |y|| = |x - y|$

In this case, the inequality $||x| - |y|| \leq |x - y|$ holds as an equality.

Case 2: $x \geq 0$ and $y \leq 0$ (or vice versa)

Assume $x \geq 0$ and $y \leq 0$. Then:

$|x| = x \quad \text{and} \quad |y| = -y$

So, the expression for $||x| - |y||$ becomes:

$||x| - |y|| = |x - (-y)| = |x + y|$

The right side, $|x - y|$, is:

$|x - y|$

Now, since $x \geq 0$ and $y \leq 0$, it is easy to see that:

$|x + y| \leq |x - y|$

Thus, the inequality $||x| - |y|| \leq |x - y|$ holds.

Case 3: $x \leq 0$ and $y \leq 0$

Assume $x \leq 0$ and $y \leq 0$. Then:

$|x| = -x \quad \text{and} \quad |y| = -y$

So:

$||x| - |y|| = |-x + y| = |y - x|$

And:

$|x - y| = |(-x) - (-y)| = |y - x|$

Thus, the inequality $||x| - |y|| \leq |x - y|$ holds as an equality in this case as well.

Case 4: $y \geq x \geq 0$ (Reverse of Case 1)

This case is symmetric to Case 1 and follows the same logic.

Conclusion

In all cases, the inequality $||x| - |y|| \leq |x - y|$ holds. This concludes the proof.

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Would you like more details on any part of the proof?

Here are some related questions you might be interested in:

1. Can you provide an example where $||x| - |y|| = |x - y|$?
2. What is the geometric interpretation of $||x| - |y|| \leq |x - y|$?
3. How does the triangle inequality compare to this inequality?
4. Is the inequality $||x| - |y|| \leq |x - y|$ strict or can it be an equality?
5. How would the proof change if $x$ and $y$ were vectors instead of real numbers?

Tip: When dealing with absolute values, breaking down the problem into cases based on the signs of the variables often simplifies the proof.