How do we prove this kind of inequality? Let $1\le p&lt;2$, 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.
Fact: For $x,y \in \mathbb{R}$ and $0 &lt; p \leq 1$ 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 $1 \leq p &lt; 2$. But nonetheless the said inequality is true even for $0 &lt; p \leq 1$ as proved above.
Fact: For $x,y \in \mathbb{R}$ and $0 &lt; p \leq 1$ 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 $1 \leq p &lt; 2$. But nonetheless the said inequality is true even for $0 &lt; p \leq 1$ as proved above.