$x^2-(m-1)x+2m-2=0\\\text{Ta có: }\Delta=(m-1)^2-4(2m-2)\\\hspace{1,9cm}=m^2-2m+1-8m+8\\\hspace{1,9cm}=m^2-10m+9\\\text{Để phương trình có 2 nghiệm }⇔\Delta≥0\\\to m^2-10m+9≥0 \to \left[\begin{array}{l}m≤1\\m≥9\end{array}\right.\\\text{Theo Viet: }\begin{cases}x_1+x_2=-\dfrac{b}{a}=m-1\\x_1x_2=\dfrac{c}{a}=2m-2\end{cases}\\\text{Theo đề bài: }x_1^2+x_2^2≥4x_1x_2\\\to (x_1+x_2)^2≥6x_1x_2\\\to (m-1)^2≥6(2m-2)\\\to m^2-2m+1≥12m-12\\\to m^2-14m+13≥0\\\to \left[\begin{array}{l}m≤1\\m≥13\end{array}\right.$
\(\Delta=\left(m-1\right)^2-8\left(m-1\right)\ge0\)
\(\Leftrightarrow\left(m-1\right)\left(m-9\right)\ge0\Rightarrow\left[{}\begin{matrix}m\le1\\m\ge9\end{matrix}\right.\)
Khi đó: \(\left\{{}\begin{matrix}x_1+x_2=m-1\\x_1x_2=2m-2\end{matrix}\right.\)
Biểu thức đề bài thật kì quặc, chắc bạn đánh nhầm
\(x_1^2+x_2^2\ge4x_1x_2\Leftrightarrow\left(x_1+x_2\right)^2\ge6x_1x_2\)
\(\Leftrightarrow\left(m-1\right)^2-12\left(m-1\right)\ge0\)
\(\Leftrightarrow\left(m-1\right)\left(m-13\right)\ge0\)
\(\Rightarrow\left[{}\begin{matrix}m\ge13\\m\le1\end{matrix}\right.\)
