Czech-Slovak-Polish Match

The result immediately follows from the following LEMMA. If $ABCD$ is any quadrilateral (convex or non-convex), then
$$(|AB| + |CD|)^2 + (|BC| + |DA|)^2 \geq 2|AC|^2 + 2|BD|^2,$$
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
$$\mathbf{a} + \mathbf{b} + \mathbf{c} + \mathbf{d} = \mathbf{0}. \qquad (1)$$
Squaring the triangle inequalities
$$|\mathbf{a}| + |\mathbf{c}| \geq |\mathbf{a} - \mathbf{c}|, \quad |\mathbf{b}| + |\mathbf{d}| \geq |\mathbf{b} - \mathbf{d}|, \qquad (2)$$
summing up and using the dot product, we get
$$\begin{aligned} (|AB| + |CD|)^2 + (|BC| + |DA|)^2 &\geq |\mathbf{a} \cdot \mathbf{c}|^2 + |\mathbf{b} - \mathbf{d}|^2 \\ &= |\mathbf{a}|^2 + |\mathbf{b}|^2 + |\mathbf{c}|^2 + |\mathbf{d}|^2 - 2 \mathbf{a} \cdot \mathbf{c} - 2 \mathbf{b} \cdot \mathbf{d} \\ &= |\mathbf{a} + \mathbf{b}|^2 - |\mathbf{c} + \mathbf{d}|^2 - 2(\mathbf{a} + \mathbf{d}) \cdot (\mathbf{b} + \mathbf{c}) \\ &= 2|\mathbf{AC}|^2 - 2 \overline{\mathbf{DB}} \cdot \overline{\mathbf{BD}} = 2|\mathbf{AC}|^2 + 2|\mathbf{BD}|^2. \end{aligned}$$
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
$$(1 - p)\mathbf{a} + (1 - q)\mathbf{b} = \mathbf{0},$$
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.