实数 $a,b,c$ 满足 $a^{2}+b^{2}+c^{2}=2023$，求 $M=\min\left\{ (a-b)^{2},(b-c)^{2},(c-a)^{2} \right\}$ 的最大值。
解：不妨设 $a\geq b\geq c$，则 $a-c\geq b-c\geq 0$
$\therefore M=\min\left\{ (a-b)^{2},(b-c)^{2} \right\}\leq(a-b)(b-c)\leq\left( \dfrac{a-b+b-c}{2} \right)^{2}=\dfrac{(a-c)^{2}}{4}$
$\because M\leq(a-b)^{2}$，$M\leq(b-c)^{2}$
$\therefore 6M\leq(a-b)^{2}+(b-c)^{2}+(a-c)^{2}\leq 3(a^{2}+b^{2}+c^{2})=2023\times 3$
$\therefore M\leq 1011.5$
不妨设 $a\geq b\geq c$，$b=c+m$，$a=c+m+n$（$m\geq 0,n\geq 0$）
$\therefore (c+m+n)^{2}+(c+m)^{2}+c^{2}=2023$
$\therefore 3c^{2}+2(2m+n)c+(m+n)^{2}+m^{2}-2023=0$
$\therefore \Delta=4(2m+n)^{2}-12\left[ (m+n)^{2}+m^{2}-2023 \right]\geq 0$
$\therefore m^{2}+mn+n^{2}\leq\dfrac{3}{2}\times 2023$
不妨设 $0\leq m\leq n$，则 $3m^{2}\leq\dfrac{3}{2}\times 2023$
且 $M=\min\left\{ (m+n)^{2},m^{2},n^{2} \right\}=m^{2}$
$\therefore M\leq 1011.5$