Product-of-combinations and Combination-of-sums Conjecture: 
$\prod _{i=1}^N \binom{n_i}{r_i} < \binom{\sum_{i=1}^N n_i}{\sum_{i=1}^N r_i}$, 
where $r_i,n_i$ are positive integers such that $\forall i, r_i \leq n_i$; and $\exists i, r_i < n_i $.
Is it true? If yes, is there any theorem and proof on it?
I have performed lots of experiments with random integers to generate the LHS and RHS and the result that no counterexamples have yet been found tend to support the conjecture.
 A: It is true indeed. Let's try to interpret the LHS and the RHS combinatorially.
LHS: You have N sets. Set $i$ has $n_i$ objects. You want to choose $r_i$ objects from the $i$th set. The LHS answers many ways can you do that.
RHS: Now imagine you grouped all the sets together. Now you have one set of $\sum n_i$ objects that you want to select $\sum r_i$ objects from.
Clearly, any selection while the objects are grouped into $N$ sets works as a selection when they're grouped together. The opposite is not true. When you group them together, you can select more than $r_i$ items from set $i$, and less from some other set to compensate. So LHS < RHS.
A: Assuming $0\le k_j\le n_j$ for all $1\le j\le m$,
$$
\begin{align}
\binom{n_1+n_2}{k_1+k_2}
&=\sum_{j=0}^{k_1+k_2}\binom{n_1}{j}\binom{n_2}{k_1+k_2-j}\tag1\\
&\ge\binom{n_1}{k_1}\binom{n_2}{k_2}\tag2
\end{align}
$$
Explanation:
$(1)$: Vandermonde Identity
$(2)$: one term ($j=k_1$) in a sum of non-negative numbers is no greater than the entire sum
Note that if $0\lt k_1+k_2\lt n_1+n_2$, there are at least two positive terms in $(1)$, so the inequality is strict.
Induction and $(2)$ show that
$$
\binom{\sum_{j=1}^mn_j}{\sum_{j=1}^mk_j}\ge\prod_{j=1}^m\binom{n_j}{k_j}\tag3
$$
and if $0\lt\sum_{j=1}^mk_j\lt\sum_{j=1}^mn_j$, the inequality in $(3)$ is strict.
