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Is $|x_1-y_2|^p+|x_2-y_1|^p \ge |x_1-y_1|^p+|x_2-y_2|^p$ for any $0 \le x_1 \le x_2$ and $0 \le y_1 \le y_2$ any positive integer $p$? I think it holds for $p=1$, since the only possible cases are

$x_1\le x_2 \le y_1 \le y_2$

$x_1\le y_1 \le y_2 \le x_2$

$x_1\le y_1 \le x_2 \le y_2$

and their symmetric ones (exchange $x$ and $y$), and we can draw on real lines to show the inequality holds for $p=1$ for all such cases.

My question is that is this true for all positive integers $p$ and how do I algebraically prove this? In addition, if this does not hold for any non-negative real number, does it hold for non-negative integers? Thank you!


My attempt (help check if my attempt is correct, thanks!):

For the last case $x_1\le y_1 \le x_2 \le y_2$, let >$|x_1-y_1|=a,|y_1-x_2|=b,|x_2-y_2|=c$, then

$|x_1-y_1|^p+|x_2-y_2|^p=a^p+c^p\le|x_1-y_2|^p+|x_2-y_1|^p=(a+b+c)^p+b^p$, where $a,b,c\ge 0$.

For the second case $x_1\le y_1 \le y_2 \le x_2$, let >$|x_1-y_1|=a,|y_1-y_2|=b,|y_2-x_2|=c$, then

$|x_1-y_1|^p+|x_2-y_2|^p=a^p+c^p\le|x_1-y_2|^p+|x_2-y_1|^p=(a+b)^p+(b+c)^p$, where $a,b,c\ge 0$.

The remaining one is the first case, which is equivalent to ask if $(a+b)^p+(b+c)^p \le (a+b+c)^p + b^p$ for any $a,b,c \ge 0$. This is true since the expansion of $(a+b+c)^p$ contains $(a+b)^p$ and $(b+c)^p$ when $p$ is a positive integer.

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  • $\begingroup$ I'm not sure this works, but my first instinct is to use the Triangle Inequality with a weird distance metric. $\endgroup$ – QuantumFool Oct 25 '16 at 17:21
  • $\begingroup$ Noting that $d_p(x,y) = |x-y|^p$ is itself a metric, can't you use the exact same argument that you did for $p=1$ to show it works for all $p$? $\endgroup$ – Nitin Oct 25 '16 at 17:43
  • $\begingroup$ @Nitin I am thinking about this, but I have trouble. We have $|x_1-y_1|^p \le |x_1-x_2|^p+|x_2-y_1|^p$ and $|x_2-y_2|^p \le |x_2-y_1|^p+|y_1-y_2|^p$. But we seem to need a lower bound for $|x_1-y_2|^p+|x_2-y_1|^p$? $\endgroup$ – Ralph B. Oct 25 '16 at 17:53
  • $\begingroup$ Looks like your proof is correct. $\endgroup$ – Nitin Oct 25 '16 at 18:31
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Your arguments are true for all integer $p$.

Here are the three cases (always $a,b,c\ge 0$):

For the last case $a^p+c^p\le(a+b+c)^p+b^p$ is true since, by expansion, already $a^p+c^p\le(a+b+c)^p$ for all $p$.

For the second case $a^p+c^p\le(a+b)^p+(b+c)^p$, which is trivially true.

For the first case we must have $(a+b)^p+(b+c)^p \le (a+b+c)^p + b^p$. Here your argument that the expansion of $(a+b+c)^p$ contains $(a+b)^p$ and $(b+c)^p$ is not true since your argument uses $b$ twice. However, the condition nevertheless holds. Consider

$$(a+b)^p+(b+c)^p = (a+b)^p+ b^p + c^p + \sum_{n=1}^{p-1} \binom{p}{n}c^n b^{p-n} $$

and $$(a+b+c)^p + b^p = (a+b)^p+ b^p + c^p + \sum_{n=1}^{p-1} \binom{p}{n}c^n (a+b)^{p-n} $$

So we need to show

$$ \sum_{n=1}^{p-1} \binom{p}{n}c^n b^{p-n} \le \sum_{n=1}^{p-1} \binom{p}{n}c^n (a+b)^{p-n} $$

which is true term-wise since always $b \le a+b$.

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