This is an idea/a partial answer.

Let $x_1,\dots,x_n$ be positive nubmers such that $\prod x_i=1$. Assume that $y_i=\prod_{k=0}^{n-1} x_{i+k}^{\alpha_k}$ for some exponents $\alpha_k$, indices are modulo $n$. Note that we may replace $\alpha_k$ to $\alpha_k+c$ for any constant $c$. Now assume that $\sum |\alpha_i+c|\leq 1$ for some $c$. Then we may choose $c$ such that $\sum |\alpha_i+c|=1$ (by continuity). Now apply Jensen inequality for the convex function $\log(1+e^x)$: Denote $\beta_k=|\alpha_k+c|$, $z_{i,k}=x_{i+k}^{{\rm sign}\,(\alpha_k+c)}$,
$$
\prod (1+y_i)=\prod_i \left(1+\prod_{k=0}^{n-1} z_{i,k}^{\beta_k}\right)\leqslant
\prod_i \prod_k (1+z_{i,k})^{\beta_k}=\prod (1+x_i),
$$   
since $\prod (1+1/x_i)=\prod_j (1+x_i)=\prod_i (1+z_{i,k})$ for any fixed $k$. 

So, if $\min_c \sum_i |\alpha_i+c|\leq 1$, we get $\prod (1+y_i)\leqslant \prod (1+x_i)$. Apply it to, say $\prod (a^3+bcd)\geqslant \prod(a^3+abc)$, here $x_1=bcd/a^3$ and $y_1=bc/a^2$. Expressing $y$'s via $x$'s is a linear system of equations with circulant matrix. Studying exponents should be more pleasant if we use some computer algebra system, which I can't, sorry.