Gọi \(M\left(0;m\right)\Rightarrow\left\{{}\begin{matrix}\overrightarrow{AM}=\left(-1;m+1\right)\\\overrightarrow{BM}=\left(-3;m-2\right)\end{matrix}\right.\)
\(T=AM^2+BM^2=1+\left(m+1\right)^2+9+\left(m-2\right)^2\)
\(=10+m^2+2m+1+m^2-4m+4\)
\(=2m^2-2m+15=2\left(m-\frac{1}{2}\right)^2+\frac{29}{2}\ge\frac{29}{2}\)
Dấu "=" xảy ra khi \(m=\frac{1}{2}\) hay \(M\left(0;\frac{1}{2}\right)\)