Expectation of a Product of Martingales Given a non-negative martingale process $(X_t)$ with its natural filtration, what can be said about
$$\mathbb{E}(X_t X_{t+1} X_{t+2} X_{t+3} | \mathcal{F}_{t-1})$$
specifically can we say 
$$\mathbb{E}(X_t X_{t+1} X_{t+2} X_{t+3} | \mathcal{F}_{t-1}) > (X_{t-1})^4$$ 
i.e. does it behave as a submartingale, without making further assumptions about covariances
 A: The answer is YES if one replaces $\gt$ by $\geqslant$ and if one assumes that $(X_t)$ has finite fourth moments. This is based on three simple properties of conditional expectation, namely:


(C) Convexity. (T) Tower property. (M) Taking out what is measurable.


(C) means that if $\varphi$ is convex and if $Z$ and $\varphi(Z)$ are integrable, then 
$$
E(\varphi(Z)|G)\geqslant \varphi(E(Z|G)).
$$
(T) means that if $H\subseteq G$ and if $Z$ is integrable, then 
$$
E(Z|H)=E(E(Z|G)|H).
$$
(M) means that if $U$ is $G$ measurable and if $UZ$ and $Z$ are integrable, then 
$$
E(UZ|G)=UE(Z|G).
$$
Now we must estimate $Y=E(Y_1|F_0)$ with $Y_1=X_1X_2X_3X_4$. 
First step 
(i) By (T) for $Z=Y_1$, $G=F_3$ and $H=F_0$, $Y=E(E(Y_1|F_3)|F_0)$. 
(ii) By (M), since $U=X_1X_2X_3$ is $F_3$ measurable, $E(Y_1|F_3)=X_1X_2X_3E(X_4|F_3)$. 
(iii) Since $E(X_4|F_3)=X_3$, $Y=E(Y_2|F_0)$ with $Y_2=X_1X_2X_3^2$.
Second step 
(i) By (T) for $Z=Y_2$, $G=F_2$ and $H=F_0$, $Y=E(E(Y_2|F_2)|F_0)$. 
(ii) By (M), since $U=X_1X_2$ is $F_2$ measurable, $E(Y_2|F_2)=X_1X_2E(X_3^2|F_2)$. 
(iii) By (C) for $\varphi:z\mapsto z^2$, $E(X_3^2|F_2)\geqslant X_2^2$.
(iv) Because every $X_i$ is nonnegative, $E(Y_2|F_2)\geqslant Y_3$ hence $Y\geqslant E(Y_3|F_0)$, with $Y_3=X_1X_2^3$. 
Third step 
Apply (T) once again, this time to $Z=Y_3$, $G=F_1$ and $H=F_0$, then (C) once again, this time with $\varphi:z\mapsto z^3$ (which is convex on $z\geqslant0$ and only there), then (M) once again with $U=X_1$, and once again the nonnegativity of every $X_i$, and you are done.
Likewise, for every $n\geqslant1$, if $(X_t)$ is integrable enough,
$$
E(X_{t+1}X_{t+2}\cdots X_{t+n}|F_t)\geqslant E((X_{t+1})^n|F_t)\geqslant (X_t)^n.
$$
The idea of synthetizing properties (C), (T) and (M) is adapted from Probability with Martingales by David Williams (see Section 9.7 Properties of conditional expectation: a list).
A: I dont know much about martingales, but I'd say yes. Calling $a,b,c,d$ your $X_t ... X_{t+3}$ conditioned to the past $F_{t-1}$, and applying the formula $E(a b) = E(a E(b|a))$ we get that the LHS is
$E(a \; E(b \;  E(c \;  E( d | a b c ))))$  (the variables in each term are assumed to be contitioned on the past values).
The most inner term is $E( d | a b c ) = c$
The next resulting term is $E( c^2 | a b ) $ . As we know that $E(c | ab) = b$, this must be greater (or equal) than $b^2$
By the same reasoning, the next term must be greater than $a^3$,  etc.
Update: Didier's answer came while I was finishing this; he says the same, basically - and better.
