**Question** : Is the following true for any $n,N\in\mathbb N$?

$$\sum_{k_1+k_2+\cdots+k_N=n,\ k_i\ge0\in\mathbb Z}\frac1{\prod_{j=1}^{N}\{(N-1)k_j+1\}}\le 1$$

**Motivation** : I've known the $N=3$ case :

$$\sum_{k_1+k_2+k_3=n,\ k_i\ge0\in\mathbb Z}\frac1{(2k_1+1)(2k_2+1)(2k_3+1)}\le 1$$

I proved this inequality by estimating the left hand side with integral. After proving this, I reached the above expectation by using computer. The above expectation seems true, but I'm facing difficulty. I would like to know how to prove this (if it's true) and any relevant references.

**Remark** : This question has been asked previously on math.SE without receiving any answers.

*Update* : I'm going to show the proof for $N=3$ case without using integral. This is because it seems that this idea can be generalized (though I'm facing difficulty).

For any non-negative integer $n$, $$\sum_{k_1+k_2+k_3=n,\ k_i\ge0\in\mathbb Z}\frac1{(2k_1+1)(2k_2+1)(2k_3+1)}\le 1$$

**Proof** : Let $A_n$ be the left hand side, and suppose that $\sum$ represents $\sum_{k_1+k_2+k_3=n,k_i\ge 0\in\mathbb Z}$. Noting that $(2k_1+1)+(2k_2+1)+(2k_3+1)=2n+3$, we get
$$\begin{align}A_n & =\sum\frac{(2k_1+1)+(2k_2+1)+(2k_3+1)}{(2n+3)(2k_1+1)(2k_2+1)(2k_3+1)}\\
& =\frac{1}{2n+3}\sum\left\{\frac{1}{(2k_1+1)(2k_2+1)}+\frac{1}{(2k_2+1)(2k_3+1)}+\frac{1}{(2k_3+1)(2k_1+1)}\right\}\\
& =\frac{3}{2n+3}\sum\frac{1}{(2k_1+1)(2k_2+1)}\\
& =\frac{3}{2n+3}\sum_{j=0}^n\sum_{k_1+k_2=j,k_i\ge 0\in\mathbb Z}\frac{1}{(2k_1+1)(2k_2+1)}\\
& \le \frac{3}{2n+3}\left(1+\frac 23 n\right)=1\end{align}$$
Here, I used
$$B_0=1, B_j\le \frac 23\ (j=1,2,\cdots,n)$$
where
$$B_j=\sum_{k_1+k_2=j,k_i\ge 0\in\mathbb Z}\frac{1}{(2k_1+1)(2k_2+1)}.$$