Can we say that $\forall x\in \Bbb R$, $g(x)>0$, $g(x)<0 $ or that $(\exists x\in \Bbb R) g(x)=0$. Let $ (a,b,c)\in \Bbb R^3 $ with $ a\ne 0 $,
$$f(x)=ax^2+bx+c$$
and
$$g(x)=f(x)+f'(x)+f''(x).$$
We assume that
$$(\forall x\in \Bbb R)\;\; f(x)>0$$
What can we conclude
$$1.   \;\;(\forall x\in \Bbb R)\;\;g(x)>0$$
$$2. \;\;(\forall x\in \Bbb R)\;\; g(x)<0$$
or
$$3. \;\; (\exists x\in \Bbb R)\;:\; g(x)=0$$
I used the fact that
$$(\forall x\in \Bbb R)\;\; f(x)>0 \implies \delta=b^2-4ac<0$$
then, i wrote $ g(x) $ as
$$g(x)=ax^2+(2a+b)x+2a+b+c$$
the discriminant is
$$\Delta=(2a+b)^2-4a(2a+b+c)$$
$$=b^2-4ac-4a^2=\delta-4a^2<0$$
i concluded that the sign of $ g(x) $ is constant  but i cannot say if it is positive or negative.
Any idea will be appreciated.

We can conclude $g(x)>0$ everywhere. If $f(x)>0$ then $a>0$, and since the coefficient of $x^2$ in $g(x)$ is also $a$, and the discriminant of $g$ is negative, we must have $g(x)>0$ for all $x$.
