\(\Leftrightarrow\left(x+\frac{1}{x}\right)^2+x+\frac{1}{x}+3m-3\ge0\) ; \(\forall x\ne0\)
Đặt \(x+\frac{1}{x}=t\Rightarrow\left|t\right|\ge2\)
\(\Rightarrow t^2+t+3m-3\ge0\) ; \(\forall\left|t\right|\ge2\)
\(\Leftrightarrow m\ge-\frac{1}{3}t^2-\frac{1}{3}t+1\)
\(\Leftrightarrow m\ge\max\limits_{\left|t\right|\ge2}f\left(t\right)\) với \(f\left(t\right)=-\frac{1}{3}t^2-\frac{1}{3}t+1\)
\(-\frac{b}{2a}=-\frac{1}{2}\notin(-\infty;-2]\cup[2;+\infty)\)
\(f\left(-2\right)=\frac{1}{3}\) ; \(f\left(2\right)=-1\Rightarrow\max\limits_{\left|t\right|\ge2}f\left(t\right)=\frac{1}{3}\)
\(\Rightarrow m\ge\frac{1}{3}\)
