$\begin{array}{l}
M \in \left( P \right) \Rightarrow M\left( {a;{a^2}} \right)\\
d\left( {M;d} \right) = \frac{{\left| {{a^2} + a + 1} \right|}}{{\sqrt 2 }} = \frac{{{{\left( {a + \frac{1}{2}} \right)}^2} + \frac{3}{4}}}{{\sqrt 2 }} \ge \frac{{3\sqrt 2 }}{8}
\end{array}$
Dấu "$=$" $...