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The problem provides a non-standard metric for vectors \( x, y \in \mathbb{R}^n \) with \( n \geq 2 \), defined by

\[
d(x, y) = \left( \sum_{k=1}^{n} |x_k - y_k|^p \right)^{1/p}
\]
for \( 0 < p < 1 \).

The task is to show that for any \( 0 < p < 1 \), there exist vectors \( x \) and \( y \) such that the triangle inequality is violated:

\[
d(x + y, 0) > d(x, 0) + d(y, 0).
\]

To break this down:

### 1. Metric Definition and Properties
The given distance function resembles the standard \( \ell_p \)-norm, except that \( 0 < p < 1 \). For these values of \( p \), this is no longer a true norm because, in this range, the function is **concave** rather than convex. This concavity implies that the distance function does not satisfy the triangle inequality.

### 2. Triangle Inequality
The triangle inequality states that for any three points \( x \), \( y \), and \( z \), the distance satisfies:

\[
d(x, z) \leq d(x, y) + d(y, z).
\]

However, since we are working with \( p \in (0,1) \), we want to show that the inequality is **reversed**, meaning that:

\[
d(x + y, 0) > d(x, 0) + d(y, 0).
\]

### 3. Strategy
We will attempt to find specific examples of vectors \( x \) and \( y \) in \( \mathbb{R}^n \) that violate the triangle inequality. To simplify, we can consider vectors with small components and focus on the behavior of the function for small distances, exploiting the concavity of the \( \ell_p \)-like metric for \( p < 1 \).

### 4. Simplified Example in \( \mathbb{R}^2 \)
Consider \( x = (1, 0) \) and \( y = (0, 1) \) in \( \mathbb{R}^2 \).

- Compute \( d(x, 0) \):

\[
d(x, 0) = \left( |1 - 0|^p + |0 - 0|^p \right)^{1/p} = \left( 1^p + 0^p \right)^{1/p} = 1.
\]

- Compute \( d(y, 0) \):

\[
d(y, 0) = \left( |0 - 0|^p + |1 - 0|^p \right)^{1/p} = \left( 0^p + 1^p \right)^{1/p} = 1.
\]

- Now, compute \( d(x + y, 0) \):

\[
x + y = (1, 1),
\]
so

\[
d(x + y, 0) = \left( |1 - 0|^p + |1 - 0|^p \right)^{1/p} = \left( 1^p + 1^p \right)^{1/p} = 2^{1/p}.
\]

### 5. Compare the Distances
We need to check if:

\[
d(x + y, 0) > d(x, 0) + d(y, 0),
\]
which translates to:

\[
2^{1/p} > 1 + 1 = 2.
\]

Since \( 2^{1/p} > 2 \) when \( 0 < p < 1 \), we see that the triangle inequality is violated for these values of \( p \).

### Conclusion
For \( 0 < p < 1 \), the function \( d(x, y) \) does not satisfy the triangle inequality. Specifically, the example above shows that for vectors like \( x = (1, 0) \) and \( y = (0, 1) \), the inequality \( d(x + y, 0) > d(x, 0) + d(y, 0) \) holds, confirming the violation.

Let me know if you'd like further details or have any questions.

Here are 5 related questions for further exploration:
1. What is the geometric interpretation of the function \( d(x, y) \) for different values of \( p \)?
2. How does the violation of the triangle inequality affect the topology induced by \( d(x, y) \)?
3. Can we construct other examples of vectors that violate the triangle inequality for different values of \( p \)?
4. What happens to the function \( d(x, y) \) as \( p \to 1 \)?
5. How does the concavity of the distance function for \( 0 < p < 1 \) affect the shape of sets in \( \mathbb{R}^n \)?

**Tip**: When exploring metrics that don't satisfy the triangle inequality, investigate the convexity or concavity of the corresponding function as it often reveals crucial insights into the behavior of the space.

For vectors x, y ∈ ℝⁿ, where n ≥ 2, define d(x, y) = ( Σ |xₖ - yₖ|ᵖ )^(1/p). Show that for every 0 < p < 1 there are vectors x and y such that d(x + y, 0) > d(x, 0) + d(y, 0). In other words, d does not satisfy the triangle inequality.

Explore the proof that for vectors in ℝⁿ and for 0 < p < 1, the distance function d(x, y) = ( Σ |xₖ - yₖ|ᵖ )^(1/p) violates the triangle inequality. This detailed step-by-step explanation illustrates how concavity for p < 1 leads to this violation in p-norms.

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