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I have been trying (unsuccessfully) for days to prove that the triangle inequality holds for this equation:

$\mathbb{R}^2$ with the metric $d((a, b), (c, d)) := (|a - c|^{3/2} + |b-d|^{3/2})^{2/3}$

Please note that this equations is not neccessarily a proper metric, in which case I'd just have to prove that it does not meet the required standards for a metric. I have been told that the p-norm triangle inequality proof would help here, but I can't see how to apply it to this instance.. Also, to clarify, the exact property I am trying to prove is that $d((a, b), (c, d)) \le d((a, b), (e, f)) + d((e, f), (c, d))$.

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user3495690
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  • well, $(a + b)^n \ge a^n + b^n$ so $(a^{3/2} + b^{3/2})^2 \ge a^3 + b^3$ so $a^{3/2} + b^{3/2} \ge (a^3 + b^3)^{1/2}$. I think that might help. – fleablood Nov 02 '16 at 22:45
  • @fleablood Does that always hold true for absolute values, though? If so, I'm having trouble seeing how I could make the substitutions to end up with the correct values.. – user3495690 Nov 03 '16 at 00:03
  • @fleablood Also, I just considered this further, what about the situation where $n = 1/2, a = 16, b = 9$? This would result in your first statement being $(16+9)^{1/2} \ge 16^{1/2} + 9^{1/2}$ implying $5 \ge 4 + 3$.. So this proof would only hold true for values of $n \ge 1$ I suppose.. – user3495690 Nov 03 '16 at 00:24
  • It was just a hint. – fleablood Nov 03 '16 at 03:05

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