Czech-Slovak-Polish Match

Without loss of generality assume that the given pentagon $ABCDE$ has the side equal to $1$. Then the length of its diagonal is equal to
$$\lambda = \frac{1 + \sqrt{5}}{2}.$$
Set $a = PA$, $b = PB$, $c = PC$, $d = PD$, $e = PE$ (Fig. 2).

Fig. 2

Applying the Ptolemy inequality for the (not necessarily convex) quadrilaterals $APDE$, $BPDC$ and $PCDE$ we obtain (respectively)
$$a + d \geqslant e\lambda, \quad b + d \geqslant c\lambda, \quad e + c \geqslant d\lambda.$$
We multiply the third inequality by $\frac{\lambda+2}{\lambda+1}$ and add together with the first and the second inequalities. As a result we obtain
$$a+b+2d+e \cdot \frac{\lambda+2}{\lambda+1} + c \cdot \frac{\lambda+2}{\lambda+1} \geq e\lambda + c\lambda + d \cdot \frac{\lambda(\lambda+2)}{\lambda+1}.$$
Grouping the respective terms, the above inequality reduces to
$$a+b \geq \frac{\lambda^2-2}{\lambda+1}(c+d+e).$$
Therefore
$$\frac{a+b}{c+d+e} \geq \frac{\lambda^2-2}{\lambda+1} = \sqrt{5}-2.$$
The equality holds if and only if the convex quadrilaterals $APDE$, $BPDC$ and $PCDE$ are cyclic. This condition is satisfied if and only if the point $P$ lies on the minor arc $AB$ of the circumcircle of the pentagon $ABCDE$ (Fig. 3). Therefore the smallest possible value of the given expression is $\sqrt{5}-2$.

Fig. 3