\(\text{Pt có 2 nghiệm}\\→Δ=[-2(m+1)]^2-4.1.(m^2+3)=4m^2+8m+4-4m^2-12=8m-8≥0\\↔8m≥8\\↔m≥1\\\text{Theo Vi-ét}: \begin{cases}x_1+x_2=2(m+1)\\x_1x_2=m^2+3\end{cases}\\S=x_1x_2+x_1+x_2\\=m^2+3+2(m+1)\\=m^2+2m+5\\=m^2+2m+1+4\\=(m+2)^2+4\ge 4\\→S_{\min}=4\\→m+2=0\\↔m=-2(KTM)\\\text{Vậy S không đạt GTNN}\)
