$\vec{AB}=(b-a;b)$
$\vec{AC}=(3-a;1)$
$\vec{CB}=(b-3;b-1)$
$AB+BC+AC=\sqrt{(a-b)^2+b^2}+\sqrt{(3-a)^2+1}+\sqrt{(3-b)^2+(b-1)^2}\ge \sqrt{(a-b+3-a+b-1)^2+(b+1+3-b)^2}=2\sqrt5$ (bdt Mincopxki)
Dấu "=" xảy ra khi $\dfrac{a-b}{b}=\dfrac{3-a}{1}=\dfrac{b-1}{3-b}\Leftrightarrow \left\{\begin{matrix}a=\dfrac52\\b=\dfrac53\end{matrix}\right.$
có gì khúc mắc em hỏi lại nhé