\(Q=\dfrac{3\left(x+1\right)}{x^3+x^2+x+1}=\dfrac{3\left(x+1\right)}{\left(x+1\right)\left(x^2+1\right)}=\dfrac{3}{x^2+1}\)

We have: \(x^2\ge0\forall x\Leftrightarrow x^2+1\ge1\forall x\)

\(\Leftrightarrow\dfrac{1}{x^2+1}\le1\forall x\)

\(\Leftrightarrow Q=\dfrac{3}{x^2+1}\le3\forall x\)

\("="\Leftrightarrow x=0\)