Two restaurants competing in the same region On a square area $[0,a]\times[0,b]$ exists a continuous function $f(x,y)$. For any $2l$ points in this area and we divide them into two sets, $K=\{k_{1},k_{2},\cdots,k_{l}\},M=\{m_{1},\cdots,m_{l}\}$.
Now define
$$M(x,y)=\begin{cases}
0&amp;\text{$d((x,y),M)&gt;d((x,y),K)$}\\
f(x,y)&amp;\text{$d((x,y),M)&lt;d((x,y),K)$}\\
\displaystyle\frac{1}{2}f(x,y)&amp;\text{$d((x,y),M)=d((x,y),K)$}
\end{cases}$$
If we arrange the set $K$ first, then $M$, can we guarantee that
$$\iint M(x,y)\text{d}\sigma\geq\frac{1}{2}\iint f(x,y)\text{d}\sigma$$
Or in a vivid way, we may regard $f(x,y)$ as the density of customers. The points are restaurants. Where to have dinner is only determined by the distance. May the initiative restaurant dominate this area or not?
Yes, one can choose $M$ such that the inequality is true.
For a given set $K=\{k_1,\dots,k_l\}$ we choose
$m_i:=k_i$.
Then, according to the definition of $M(x,y)$, we have
$M(x,y)=\frac12 f(x,y)$ for all $(x,y)$.
This implies the inequality (in this situation, it is even an equality).
Yes, one can choose $M$ such that the inequality is true.
For a given set $K=\{k_1,\dots,k_l\}$ we choose
$m_i:=k_i$.
Then, according to the definition of $M(x,y)$, we have
$M(x,y)=\frac12 f(x,y)$ for all $(x,y)$.
This implies the inequality (in this situation, it is even an equality).