find the sign of $f$ from a given inequallity

Question

I have a function $f:\mathbb{R}\rightarrow\mathbb{R}$ which can derives twice for which is true that: $$f''(x)f(x)-(f'(x))^2>0\text{ }\forall x\in\mathbb{R}$$ And I want to prove that $f(x)>0\text{ }\forall x\in\mathbb{R}$. So: $$f''(x)f(x)-(f'(x))^2>0\iff f''(x)f(x)>(f'(x))^2\ge 0\iff f''(x)f(x)>0\text{ }\forall x\in\mathbb{R}$$ I am thinking now of dividing the give relationship with $f''(x)f(x)>0\text{ }\forall x\in\mathbb{R}$, but am I on the right way? Any ideas?

Answer 1

Notice that if a function $f : \mathbb{R} \rightarrow \mathbb{R}$ satisfies your inequality, then so does $-f$.
Therefore, one can guess that the result is false - actually, one has to check that such a function $f$ exists.

For example, consider the function $f : x \mapsto -e^{e^{x}}$.

Then it is obvious that for all $x \in \mathbb{R}$, $f(x) < 0$.

However, for all $x \in \mathbb{R}$, we have $$f'(x) = -e^{x}. e^{e^{x}} \text{ and } f''(x) = -e^{x}. e^{e^{x}} - \left(e^{x}\right)^{2}. e^{e^{x}}$$ and thus, $$f''(x). f(x) -\left(f'(x)\right)^{2} = e^{x}. \left(e^{e^{x}}\right)^{2} > 0$$ which contradicts your statement.