I need help in deducing an inequality related to factorial.
Information -> n, r j, a all belongs to positive integers and j belongs to set { 0,1,...,n}
It is to be proved that $\frac { ( rn+j+1)! } { ( j+1)! {(j! × ( n-j)!}^r}$ × $\frac { (r+1)n -j+ 1)! } { ( n-j+1)!{(j! ×(n-j)!)}^r}$ × $\frac{{ n! }^{a-2r}}{ (j! ( n-j)!)^{a-2r} } $$ { j(n-j)}^a $ $ 8^a$ $ \leq$ $ (2r+1)^{(2r+1)n+2} × 2^{(a-2r) n } × (2n^2)^a$
Authors just gave this expanation $\frac { ( rn+j+1)! } { ( j+1)! {(j! × ( n-j)!}^r}$ and $\frac { (r+1)n -j+ 1)! } { ( n-j+1)!{(j! ×(n-j)!)}^r}$ are increased respectively by $ { (2r+1) }^{ rn+j+1} $ and $ { ( 2r+1) }^ {( r+1)n-j +1}$ .
Using this explanation one can easily obtain this part ${2r+1} ^{(2r+1) n+2}$ on the RHS of inequalitybut I don't know how to obtain complete inequality. Despite thinking a lot I am unable to understand it.
Please help.
$\frac{{ n! }^{a-2r}}{(j! ( n-j)!)^{a-2r}} = \binom{n}{j}^{a-2r}$. Since $\binom{n}{j}\leq 2^n$, this term is bounded by $2^{n(a-2r)}$. Since $j(n-j)\leq n^2/4$, $8j(n-j) \leq 2n^2$ and thus $(j(n-j))^a8^a \leq (2n^2)^a$. Combining these bounds proves the last part of the inequality.
Update: Regarding the first two terms, they follow from the multinomial theorem: For example, the first term of the LHS of the inequality is one of the terms in the multinomial expansion of $(1 + \ldots + 1)^{j+1 + rj + r(n-j)} = (2r+1)^{rn+j+1}$