Inspired by parv mor

I think that there is some gaps in my proof, but: Divide the real line into segements of the form $(a,b)$ when in the segments, $f''(x)$ does not change sign, and such that in a segment $(b,b+\epsilon$ ) $f''(x)$ change sign, and the same about a neighborhood of $(a-\epsilon,a)$ . Now, without loss of generality, assume that in the segment $\sigma=(a,b)$ we have that for each $x\in\sigma, f''(x)>0$ . To satisfy the inequality, we must have $f(x)<0$ if $x\in\sigma$ . But in $\sigma, f(x)$ is convex and thus $f(a)<0, f(b)<0$ . But in the segment $(b,\infty)$ we a segment of the form of $\sigma$ , call it $\sigma_1$ , in which $f''(x)<0$ and with its leftmost point as b ( it is of the form $(b,a_1)$ with $a_1$ not nesecerally finite). Thus in a right neighborhood $B$ of b which is contained in $\sigma_1$ we must have that $f(x)>0, x\in B$ , which contradicts the fact that $f(x)$ is contniuous.

Let, $f(x)$ = $ax^2+bx+c$ . Then we assume that for every $x$ , $f(x) > 0$ and $f''(x) < 0$ .

for $f(x) > 0$ ,

$x>\frac{√(b^2-4ac)-b}{2a}$ and $x<\frac{b-√(b^2-4ac)}{2a}$

As, $f''(x) < 0$ , so $a<0$ . So for one fixed value of $a$ , $f''(x)$ is either $>0$ or $<0$ , but for $f(x)$ ,there is always a range of $x$ in which it is increasing or decreaing. So $f(x)*f''(x) < 0$ is not possible for every $x$