Let $a,b\in{\mathbb Z}^{\ge0}$ and $h\in{\mathbb R}(d_1,d_2)$ be such that $$a\ge b, \quad h(d_1,d_2)>0~~\forall\,d_1,d_2\in{\mathbb Z}^+, \quad \lim_{d_1,d_2\longrightarrow\infty}\frac{h(d_1,d_2)}{d_1^ad_2^a/(d_1\!+\!d_2)^b}\in{\mathbb R}^+\,.$$ The limit statement means that these fractions are within $\epsilon$ of some number for all $d_1,d_2\!\ge\!d(\epsilon)$.

For $d\!=\!1,2,\ldots$, define the numbers $n_d\!\in\!{\mathbb R}^+$ inductively by $$n_1=1, \qquad n_d=\sum_{\begin{subarray}{c}d_1+d_2=d\\ d_1,d_2\ge1\end{subarray}}\!\!\!\! h(d_1,d_2)n_{d_1}n_{d_2} \quad\forall~d>1.$$ Are the numbers $\sqrt[d]{n_d}$ eventually increasing? In other words, is there $d^*\!\in\!{\mathbb Z}^+$ such that $$\sqrt[d]{n_d}\le\sqrt[d+1]{n_{d+1}} \qquad\forall~ d\ge d^*.$$ This is true if $h(d_1,d_2)=d_1^a d_2^a/(d_1\!+\!d_2)^a$. In general, these roots are bounded above and below away from zero.

a prioriupper bounds on $h(d_1,d_2)$ for $d_1+d_2 < 2d(\epsilon)$? If so, you could try to make my previous remark into an induction proof. $\endgroup$ – Jason Starr Mar 28 '16 at 16:57