Ta có: [tex]m=\frac{\sqrt{x^2-x+1}+\sqrt{x^2+x+1}}{\sqrt{\frac{3x^2+4x+8}{x^2+x+2}}}[/tex]
Ta thấy: [tex]\sqrt{x^2-x+1}+\sqrt{x^2+x+1}=\sqrt{(\frac{1}{2}-x)^2+\frac{3}{4}}+\sqrt{(x+\frac{1}{2})^2+\frac{3}{4}} \geq \sqrt{(\frac{1}{2}-x+x+\frac{1}{2})^2+(\frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{2})^2}=2[/tex]
[tex]\sqrt{\frac{3x^2+4x+8}{x^2+x+2}}=\sqrt{4-\frac{x^2}{x^2+x+2}}\leq 2\Rightarrow m\geq \frac{2}{2}=1[/tex]
Vậy Min m = 1.