Question: The question is in the title, really. More precisely, suppose $F:[0,1] \to \mathbb{R}$ is continuously differentiable and satisfies 

(i) $F(t)F'(t) \le 0$ for all $t \in [0,1]$

(ii) $F(0) = 1$

Does this imply that $F(t) \geq 0$ for all $t \in [0,1]$?

My intuition is that (i) implies that $F$ and $F'$ have different signs. Since $F(0) = 1$, this means that $F$ must cross the $t$-axis at some point, but the instant it becomes negative it would have to immediately shoot up again. I think this is not possible, but I have not been able to come up with a proof or counter-example. Thanks for your help!

Answer: 
Condition $(i)$ implies that $t \mapsto F(t)^2$ is decreasing. Since $F(0)=1$, we have two cases: $F(t)>0$ for all $t \in [0,1]$ and we are done, or there exists $t\_0 \in [0,1]$ with $t\_0=\inf \{ t \in [0,1] : f(t)=0\}$. But then $F^2(t\_0)=0$ and since $F^2$ is decreasing, it must follow that $0\leq F(t)^2\leq F(t\_0)=0,\ \forall t \in [t\_0,1]$, concluding that $F(t)=0,\ \forall t \geq t\_0$. Since before $t\_0$ the function $F$ is positive, we are done.
Your reasoning is sound and can be made rigorous like this: Assume that $F(t)<0$ for some $t>0$. Then by the intermediate value theorem there must be an $x$ in $(0,t)$ such that $F(x)=0$. Let $u$ be the supremum of all such $x$'s. By continuity of $F$, we must have $F(u)=0$. Apply the mean value theorem to the interval $[u,t]$. This gives some $y$ in $(u,t)$ such that $F'(y) = \frac{F(t)}{t-u}$ which is negative. However, since $y>u$ we must have $F(y)<0$, contradicting $F(y)F'(y) \le 0$.