Let $P(x)$ be a real polynomial with a positive leading coefficient, and $k\geq 2$ an integer.
Suppose that $Q(x)=P(P(\dots(P(x))\dots))$, where there are $k$ iterations of $P$'s, has at least one positive root and no real nonpositive root. Does it always hold that $P(x)$ has at least one positive root and no real nonpositive root?
It is true that $P(x)$ must have at least one positive root - if not, $P(x)>0$ for all $x>0$, so $P(P(x))>0$ for all $x>0$ and so on, implying that $Q(x)$ has no positive root either. The question is now whether $P(x)$ cannot have a real nonpositive root.
Let $P_k(x)=P(P(...(x))$ with $k$ iterations of $P$.
Suppose (for contradiction) that $P(x)$ has a non-positive root $s$ and let $r$ be a positive root of $P(x)$ so that $P(x)=(x-r)(x-s)f(x)$.
Then $P_2(x)= (P(x)-r)(P(x)-s)g(x)$. Note that $P(x)$ must have arbitarily large negative values or arbit. large positive values to the left of $s$; thus $P(x)=r$ or $P(x)=s$ for some $x \leq s \leq 0$.
Now $P_3(x)=(P_2(x)-r)(P_2(x)-s)h(x)$ and repeating the same argument, we inductively prove that each $P_k(x)$, and hence $Q(x)$, must have both a positive root and a non-positive root, giving the desired contradiction.