Suppose that f (x) is a quadratic expression positive for all real x. If $\mathrm{g}\left(\mathrm{x}\right)=\mathrm{f}\left(\mathrm{x}\right)+{\mathrm{f}}^{\mathrm{\prime }}\left(\mathrm{x}\right)+{\mathrm{f}}^{\prime \prime }\left(\mathrm{x}\right)$, then for any real x (where f'(x) and f"(x) represent 1st and 2nd derivative, respectively)

Let $\mathrm{f}\left(\mathrm{x}\right)={\mathrm{ax}}^2+\mathrm{bx}+\mathrm{c}$ be a quadratic expression such that

$\mathrm{f}\left(\mathrm{x}\right)>0\text{ for all }\mathrm{x}\in \mathrm{R}\text{. Then, }\mathrm{a}>0\text{ and }{\mathrm{b}}^2-4\mathrm{ac}<0\text{. Now, }$

$\begin{array}{r}\mathrm{g}\left(\mathrm{x}\right)=\mathrm{f}\left(\mathrm{x}\right)+{\mathrm{f}}^{\mathrm{\prime }}\left(\mathrm{x}\right)+{\mathrm{f}}^{\prime \prime }\left(\mathrm{x}\right)\\ \Rightarrow \mathrm{g}\left(\mathrm{x}\right)={\mathrm{ax}}^2+\mathrm{x}(\mathrm{b}+2\mathrm{a})+(\mathrm{b}+2\mathrm{a}+\mathrm{c})\end{array}$

Discriminant of g(x) is

$\begin{array}{l}\mathrm{D}=(\mathrm{b}+2\mathrm{a}{)}^2-4\mathrm{a}(\mathrm{b}+2\mathrm{a}+\mathrm{c})\\ ={\mathrm{b}}^2-4{\mathrm{a}}^2-4\mathrm{ac}\\ =\left({\mathrm{b}}^2-4\mathrm{ac}\right)-4{\mathrm{a}}^2\\ <0\end{array}$

$\left(\because {\mathrm{b}}^2-4\mathrm{ac}<0\right)$

Therefore, g(x) > 0 for all $\mathrm{x}\in \mathrm{R}$.