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Let $1\le p<2$, we want to show that for any $a,b,c,d\in\Bbb R$ the following holds: $$ |a-b|^p + |b-c|^p + |c-d|^p + |d-a|^p \ge |a-c|^p + |b-d|^p $$ The equation is symmetric in $a,c$ and $b,d$.

Since I am quite bad at solving this kind of inequality in general, I would really love you could explain the thought process behind solving inequalities like this one.

PS. The tag functional analysis is because I encountered this in a context related to $L^p$ spaces.

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    To simplify the problem, you may assume that $a+c = 0$ and that either also $b + d = 0$ or (after rescaling) $b + d = 2$. The inequality therefore may be rewritten in terms of just two variables. – 2017-02-20
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    @HansEngler I am so sorry but I must admit that I cannot see why we can assume $a+c=0$. Solving inequality has never been my strongest point. – 2017-02-20
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    Subtract the constant $(a+c)/2$ from all variables in the problem. – 2017-02-20
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    I am probably blind: How is this the triangle inequality for $p=1$? – 2017-02-20
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    @MartinR The case $p=1$ follows from the triangle inequality by adding up these two: $$\left(|a-b| + |b-c|\right) + \left(|c-d| + |d-a|\right) \ge |a-c| + |a-c| = 2\,|a-c|$$ $$\left(|a-b| + |d-a|\right) + \left(|b-c| + |c-d|\right) \ge |b-d| + |b-d| = 2\,|b-d|$$ – 2017-02-21
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    FWIW the case $p=2$ follows from the identity: $$(a-b)^2+(b-c)^2+(c-d)^2+(d-a)^2-(a-c)^2-(b-d)^2 = (a-b+c-d)^2$$ (which is related to the generalization of the [parallelogram law](https://en.wikipedia.org/wiki/Parallelogram_law) for convex quadrilaterals). – 2017-02-21

1 Answers 1

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Fact: For $x,y \in \mathbb{R}$ and $0 < p \leq 1$ we have $(|x+y|)^p\le|x|^p+|y|^p.$

For proof: $L^p$ norm and triangle inequality

Using the fact above you get the following 4 inequalities,

$$(|a-c|)^p\le|a-b|^p+|b-c|^p,$$ $$(|a-c|)^p\le|c-d|^p+|d-a|^p,$$ $$(|b-d|)^p\le|b-c|^p+|c-d|^p,$$ $$(|b-d|)^p\le|a-b|^p+|d-a|^p.$$

Now just combine all the above inequalities to get the result that you are after.

Have to work out the proof of the inequality when $1 \leq p < 2$. But nonetheless the said inequality is true even for $0 < p \leq 1$ as proved above.

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    So, for instance, if $x=y=1$ and $p = \frac 32$ you have a nice proof of the fact that $2^{3/2} \le 2$? – 2017-02-20
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    You should really look up the _true_ definition of $l_p$ norm... – 2017-02-20
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    @UmbertoP. Thanks for the comment. The fact was true for $0 < p \leq 1$, sorry for the confusion. – 2017-02-21