The answer is NO:
Suppose the statement is true. Then f(x)>0, f''(x)<0 for all x>0. Moreover, \lim_{x\rightarrow +\infty} f''(x)<0 if it exists.
Let f(0)>0, f'(x) is monotonically decreasing, then \lim_{x\rightarrow +\infty} f'(x) is either -\infty or a constant.
If \lim_{x\rightarrow +\infty} f'(x)=-\infty, then f is monotonically decreasing on [T,\,+\infty) for some T large enough and \lim_{x\rightarrow+\infty} f(x) must be negative. Otherwise, \lim_{x\rightarrow+\infty} f(x)=c_0\geq 0 and there exists a unbounded sequence \{x_n\} so that \lim_{n\rightarrow +\infty} f'(x_n)=0. Contradiction;
if \lim_{x\rightarrow +\infty} f'(x)=c, where c is a constant, then there exists a unbounded sequence \{x_n\} so that \lim_{n\rightarrow +\infty} f''(x_n)=0, but we already have \lim_{n\rightarrow +\infty} f''(x_n)<0. Contradiction.
P. S.
Note that the assumption is f(x)\cdot f''(x)\leq -1 on [0,\,+\infty). If we change it into f(x)\cdot f''(x)<0, then we cannot have \lim_{x\rightarrow +\infty} f''(x)<0 and in this case the statement is true:
e.g.: Let f(x)=\ln (x+2). Then we have
\begin{align*}<br />
f'(x)&=1/(x+2),\\<br />
f''(x)&=-1/(x+2)^2.<br />
\end{align*}
Therefore, for all x\geq 0, we have
\begin{align*}f(x)\cdot f''(x)&=-\frac{\ln (x+2)}{(2+x)^2}<0 \\<br />
\intertext{and}<br />
f(x)&>0.<br />
\end{align*}
