If $f$ is a quadratic and $f(x)>0\;\forall x$, and $g= f + f' + f''$, prove $g(x)>0\; \forall x$ If $f(x)$ is a quadratic expression such that $f(x)>0\;\forall x\in\mathbb{R},$ and if $g(x)=f(x)+f'(x)+f''(x),$
Then prove that $g(x)>0\; \forall \; x\in \mathbb{R}$.
$\bf{My\; Trial \; Solution::}$ If $f(x)>0\;\forall x\in \mathbb{R}$. Then function $f(x)$ has Minimum value .
Let Minimum occur at $x=x_{0}$. Then $f(x)_{Min} = f(x_{0})>0$
Now Given $g(x) = f(x)+f'(x)+f''(x)$. Then $g'(x) = f'(x)+f''(x)+f'''(x).$
Now at $x=x_{0}\;\;,$ Value of $g(x_{0}) = f'(x_{0})+f''(x_{0})+f'''(x_{0}) = 0+f''(x_{0})+f'''(x_{0})$
Now I did not understand How can i solve after that, 
Help me 
Thanks

Suppose $f(x)=x^2+ax+b$ with $b=f(0)>0$ and $a^2<4b$. Then,
$$
g(x)=x^2+(a+2)x+(2+a+b).
$$
We note that $g(0)=2+a+b=1+f(1)>0$ and
$$
(a+2)^2-4(2+a+b)=a^2+4a+4-8-4a-4b=(a^2-4b)-4<0
$$
so $g$ is never $0$ for real $x$. You now can infer that $g$ is always positive.
The more general case $f(x)=C(x^2+ax+b)$ is the same. It just adds the constant $C>0$ to everything.
