\(N\left(1;n\right)\)
\(NA+NB=\sqrt{2^2+\left(2-n\right)^2}+\sqrt{1^2+\left(n+1\right)^2}\)
Bất đẳng thức Minkowski:
\(\sqrt{2^2+\left(2-n\right)^2}+\sqrt{1^2+\left(n+1\right)^2}\ge\sqrt{\left(2+1\right)^2+\left(2-n+n+1\right)^2}=3\sqrt{2}\)
Dấu "=" xảy ra khi \(\dfrac{2-n}{2}=\dfrac{n+1}{1}\Leftrightarrow n=0\)
Vậy \(N\left(1;0\right)\).