Let $\displaystyle f: \mathbb{R} \rightarrow \mathbb{R}$ have a third derivative that exists at all points. Show that there must exist at least one point $\displaystyle \xi$ for which
$\displaystyle f(\xi)f^{'}(\xi)f^{''}(\xi)f^{'''}(\xi) \geq 0$
Let $\displaystyle f: \mathbb{R} \rightarrow \mathbb{R}$ have a third derivative that exists at all points. Show that there must exist at least one point $\displaystyle \xi$ for which
$\displaystyle f(\xi)f^{'}(\xi)f^{''}(\xi)f^{'''}(\xi) \geq 0$
Hmm
Restrict the function to $\displaystyle f:[a,b] \rightarrow \mathbb{R}$, where $\displaystyle f, f^{'}, f^{''}$ are increasing in that interval and $\displaystyle f(x) \geq 0$ for $\displaystyle x \in [a,b]$. Since $\displaystyle f$ is continuous, by MVT, there exists $\displaystyle c \in [a,b]$ such that $\displaystyle \frac{f(b)-f(a)}{b-a} = f^{'}(c)$. Then $\displaystyle f^{'}(c) \geq 0$. Since $\displaystyle f^{'}$ is continuous, by MVT, there exists $\displaystyle d \in [c,b]$ such that $\displaystyle \frac{f^{'}(b)-f^{'}(c)}{b-c} = f^{''}(d)$. Then $\displaystyle f^{''}(d) \geq 0$. Since $\displaystyle f^{''}$ is continuous, by MVT, there exists $\displaystyle \xi \in [d,b]$ such that $\displaystyle \frac{f^{''}(b)-f^{''}(d)}{b-d} = f^{'''}(\xi)$. Then $\displaystyle f^{'''}(\xi) \geq 0$. Since $\displaystyle f, f^{'}, f^{''}$ are increasing in that interval, therefore $\displaystyle f^{'}(\xi) \geq f^{'}(c)$ and $\displaystyle f^{''}(\xi) \geq f^{''}(d)$. These imply that $\displaystyle f(\xi), f^{'}(\xi), f^{''}(\xi), f^{'''}(\xi) \geq 0$. Hence the conclusion holds.
This could be one possible case out of so many different cases. Anyway, is this particular proof correct?
Please advise. Thank you.
Here are some vague thoughts, which you may be able to convert into a proof.
First, if any one of the four functions $\displaystyle f,\ f',\ f'',\ f'''$ vanishes at some point $\displaystyle \xi$, then $\displaystyle f(\xi)f'(\xi)f''(\xi)f'''(\xi) = 0$ and so the result is proved. It follows from the intermediate value theorem that each of the (continuous) functions $\displaystyle f,\ f',\ f''$ always has the same sign. By Darboux's theorem the same is true for $\displaystyle f'''$.
Replacing $\displaystyle f$ by $\displaystyle -f$ if necessary, you can assume that $\displaystyle f$ is always positive
Now suppose that $\displaystyle f'$ is always negative. By considering what happens as $\displaystyle \xi\to\infty$, show that $\displaystyle f''$ must be positive. The same argument, repeated, will show that $\displaystyle f'''$ is negative. Therefore the product $\displaystyle ff'f''f'''$ must be positive.
By considering what happens as $\displaystyle \xi\to-\infty$, use the same ideas to show that if $\displaystyle f'$ is positive then all four functions must be positive, hence so is their product.