**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](https://math.stackexchange.com/questions/520220/sum-k-1k-2-cdotsk-n-n-k-i-ge0-in-mathbb-z-frac1-prod-j-1n-n-1k) 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
$$B\_j=\sum\_{k\_1+k\_2=j,k\_i\ge 0\in\mathbb Z}\frac{1}{(2k\_1+1)(2k\_2+1)}.$$