A college math problem

Is there a positive and twice differentiable function f defned on [0,∞) such
that f(x)f’’(x) ≤ -1 on [0,∞) ? Why or why not???

The answer is NO:

Suppose the statement is true. Then $f(x)>0, f''(x)<0 $ for all [tex] x>0 [/tex]. Moreover, $\lim_{x\rightarrow +\infty} f''(x)<0 $ if it exists. Let [tex] f(0)>0, f'(x) [/tex] is monotonically decreasing, then [tex]\lim_{x\rightarrow +\infty} f'(x) [/tex] is either [tex] -\infty [/tex] or a constant.

If [tex]\lim_{x\rightarrow +\infty} f'(x)=-\infty [/tex], then [tex]f [/tex] is monotonically decreasing on [tex] [T,\,+\infty)[/tex] for some [tex]T [/tex] large enough and [tex]\lim_{x\rightarrow+\infty} f(x) [/tex] must be negative. Otherwise, [tex]\lim_{x\rightarrow+\infty} f(x)=c_0\geq 0 [/tex] and there exists a unbounded sequence [tex]\{x_n\}[/tex] so that [tex]\lim_{n\rightarrow +\infty} f'(x_n)=0 [/tex]. Contradiction;

if [tex]\lim_{x\rightarrow +\infty} f'(x)=c [/tex], where [tex]c[/tex] is a constant, then there exists a unbounded sequence [tex]\{x_n\}[/tex] so that [tex]\lim_{n\rightarrow +\infty} f''(x_n)=0 [/tex], but we already have [tex]\lim_{n\rightarrow +\infty} f''(x_n)<0 [/tex]. Contradiction. P. S. Note that the assumption is [tex]f(x)\cdot f''(x)\leq -1 [/tex] on [tex][0,\,+\infty)[/tex]. If we change it into [tex]f(x)\cdot f''(x)<0 [/tex], then we cannot have [tex]\lim_{x\rightarrow +\infty} f''(x)<0 [/tex] and in this case the statement is true: e.g.: Let [tex]f(x)=\ln (x+2)[/tex]. Then we have [LEFT] [tex]\begin{align*} f'(x)&=1/(x+2),\\ f''(x)&=-1/(x+2)^2. \end{align*}[/tex] [/LEFT] Therefore, for all [tex]x\geq 0[/tex], we have [LEFT][tex]\begin{align*}f(x)\cdot f''(x)&=-\frac{\ln (x+2)}{(2+x)^2}<0 \\ \intertext{and} f(x)&>0.
\end{align*}[/tex][/LEFT][/QUOTE]