Originally Posted by
tonio That's a parabola , so it'll be always positive iff (1) it is a "smiling" or convex upwards parabola (i.e., its higher coefficient is positive), and (2) its discriminant is negative, i.e.: iff it has no real roots, so:
$\displaystyle 2x^2 +6x +1 +k(x^2+2)\Longrightarrow (k+2)x^2+6x+(2k+1)>0\,\,\forall\,x\in \mathbb{R}\Longleftrightarrow$ $\displaystyle \,(1)\,\,k+2>0\,\,and\,\,(2)\,\,\Delta=b^2-4ac=6^2-4(k+2)(2k+1)<0$:
$\displaystyle (1)\,\,k+2>0\,\Longrightarrow\,k>-2\,;$
$\displaystyle (2)\,\,\Delta=36-8k^2-20k-8<0\,\Longrightarrow$ $\displaystyle \,-8k^2-20k+28<0\,\,(divide\,\,by\,\,-4)\,\,\Longrightarrow\,2k^2+5k-7>0$ $\displaystyle \Longrightarrow\,(2k+7)(k-1)>0\,\Longrightarrow\,k<-\frac{7}{2}\,\,or\,\,k>1$
Taking together both conditions for k on (1)-(2) the solution is immediate.
Tonio