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

Without loss of generality, we can assume that $f(x)= (x-a)^2 + b^2$. 
Note that $g$ is also a quadratic polynomial. We find the critical point of $g$. 
$$
g'(x) = f'(x) +  f''(x)$$
Note also that third derivative of $f$ is zero.
Then we see that the critical point is $x=a-1$, and also it gives the minimum value of $g$. 
Plug in the value to $g$, then we have
$$
g(a-1) = f(a-1) + g'(a-1) = f( a-1) > 0.$$
Therefore $g(x)>0$ for all $x\in \mathbb{R}$. 
