Czech-Slovak-Polish Match

The result immediately follows from the following LEMMA. If $ABCD$ is any quadrilateral (convex or non-convex), then $$(|AB|+|CD|{)}^2+(|BC|+|DA|{)}^2\ge 2|AC{|}^2+2|BD{|}^2,$$ with the equality if and only if $ABCD$ is a parallelogram.

PROOF OF LEMMA. The vectors $\mathbf{a}=\overrightarrow{AB}$, $\mathbf{b}=\overrightarrow{BC}$, $\mathbf{c}=\overrightarrow{CD}$ and $\mathbf{d}=\overrightarrow{DA}$ obviously satisfy $$\mathbf{a}+\mathbf{b}+\mathbf{c}+\mathbf{d}=0.(1)$$ Squaring the triangle inequalities $$|\mathbf{a}|+|\mathbf{c}|\ge |\mathbf{a}-\mathbf{c}|,|\mathbf{b}|+|\mathbf{d}|\ge |\mathbf{b}-\mathbf{d}|,(2)$$ summing up and using the dot product, we get $$\begin{array}{rl}(|AB|+|CD|{)}^2+(|BC|+|DA|{)}^2& \ge |\mathbf{a}\cdot \mathbf{c}{|}^2+|\mathbf{b}-\mathbf{d}{|}^2\\ & =|\mathbf{a}{|}^2+|\mathbf{b}{|}^2+|\mathbf{c}{|}^2+|\mathbf{d}{|}^2-2\mathbf{a}\cdot \mathbf{c}-2\mathbf{b}\cdot \mathbf{d}\\ & =|\mathbf{a}+\mathbf{b}{|}^2-|\mathbf{c}+\mathbf{d}{|}^2-2(\mathbf{a}+\mathbf{d})\cdot (\mathbf{b}+\mathbf{c})\\ & =2|\mathbf{AC}{|}^2-2\overline{\mathbf{DB}}\cdot \overline{\mathbf{BD}}=2|\mathbf{AC}{|}^2+2|\mathbf{BD}{|}^2.\end{array}$$ Thus the desired inequality is proven. If it is an equality, then (2) must become equalities as well, hence $\mathbf{c}=-p\mathbf{a}$ and $\mathbf{d}=-q\mathbf{b}$ for some positive real $p$ and $q$. Substituting this into (1) yields $$(1-p)\mathbf{a}+(1-q)\mathbf{b}=0,$$ hence both $1-p$ and $1-q$ vanish. Thus $\mathbf{c}=-\mathbf{a}$ (and $\mathbf{d}=-\mathbf{b}$) which means that $ABCD$ is a parallelogram.

Conversely, if $ABCD$ is a parallelogram, then $|AB|=|CD|$, $|BC|=|DA|$ and thus the proved inequality verges into the well-known parallelogram equality, which itself follows from our solution if we take (2) as two obvious equalities.