In the plane there are given vectors $a, b, c, d, $, the sum of which is equal to $0$. Prove the inequality $$\mid a \mid+ \mid b \mid + \mid c \mid + \mid d \mid \geq \mid a+d \mid + \mid b+d \mid + \mid c+d \mid.$$
Solution: The right side of the inequality, like the left side, is symmetric with respect to $a, b, c$ and $d$ (under the condition that their sum is equal to $0$); since $b+c= - \left(a+d\right)$ etc., it is equal to half the sum of the lengths of all pairwise sums of these vectors.
Given this, for the given $4$ vectors with sum $0$, we can choose the notation so that the polyline consisting of the vectors $$\vec{AB}=a, \vec{BC}=b, \vec{CD }=c, \vec{DA}=d,$$ is self-intersecting. For this purpose, it is enough that the vectors $a, b, c$, represented from one point $O$, lie in the same half-plane, where $a$ and $c$ are on the same side from the line passing through the point $O$ and parallel to $b$. Then $$\mid a+d \mid + \mid c+d \mid=BD+AC \leq \mid b \mid + \mid d \mid,$$ and it remains to add $$\mid b+d\mid = \mid a+c \mid \leq \mid a \mid + \mid c \mid.$$
How can we explain this step $$BD+AC \leq \mid b \mid + \mid d \mid$$ in more detail? Since this inequality can be quite non-obvious.. Any hint would help a lot! Thanks!
Since our vectors are parallel to some plane, we need to prove that: $$|a|+|b|+|c|+|a+b+c|\geq|a+b|+|a+c|+|b+c|,$$ where $a$, $b$ and $c$ are complex numbers and since $$|a|^2+|b|^2+|c|^2+|a+b+c|^2=|a+b|^2+|a+c|^2+|b+c|^2,$$ it's enough to prove that: $$\sum_{cyc}(|ab|+|c(a+b+c)|)\geq\sum_{cyc}|(a+c)(b+c)|,$$ which is true by the triangle inequality.