Show that $\frac{1}{N} \sum_{i=1}^N \frac{n_i - 1}{n_i + 1} \leq \frac{\left(\prod_{i=1}^N n_i \right) - 1}{\left(\prod_{i=1}^N n_i \right) + 1}$

Question

I'm not a mathematician but I came across the inequality $\frac{1}{N} \sum_{i=1}^N \frac{n_i - 1}{n_i + 1} \leq \frac{\left(\prod_{i=1}^N n_i \right) - 1}{\left(\prod_{i=1}^N n_i \right) + 1}$ in some work I am looking at. We have that $n_i \in \mathbb{N}, n_i \geq 2$ for all $i$.

I thought it would be reasonably easy to show by induction, so I started looking at how much each side increases by when going from $N=k$ to $N=k+1$, hoping to show that the increase of the LHS is smaller than that of the RHS so you would still know the inequality holds, but this has been fruitless so far. Is there some easy way to show this? I've verified it numerically and it seems to hold for any combination of $N$ and $n_i$'s I throw at it, but I can't seem to prove it. Does anybody know the best way to go about showing this?

Answer 1

You can prove this using the Jensen inequality.

The function $f(x) = \dfrac{x-1}{x+1}$ is concave (and increasing in $\mathbb{R}_+$). Hence by Jensen inequality, you have : $$\dfrac{1}{N}\sum_{k=1}^N f(n_k) \leq f\left(\dfrac{1}{N}\sum_{k=1}^Na_k\right)$$ And since $f$ is also increasing, you have $$ f\left(\dfrac{1}{N}\sum_{k=1}^Na_k\right) \leq f\left(\prod_{k=1}^Na_k\right)$$ We have proven that : $$ \dfrac{1}{N}\sum_{k=1}^N f(n_k) \leq f\left(\prod_{k=1}^Na_k\right) \square$$

Answer 2

Another way.

Since for naturals $n_i$ we have $$\frac{n_1+n_2+...+n_{N}}{N}\leq\frac{\prod\limits_{i=1}^nn_i+\prod\limits_{i=1}^nn_i+...+\prod\limits_{i=1}^nn_i}{N}=\prod\limits_{i=1}^nn_i,$$ by C-S we obtain: $$\frac{1}{N}\sum_{i=1}^N\frac{n_i-1}{n_i+1}=\frac{1}{N}\left(\sum_{i=1}^N\left(\frac{n_i-1}{n_i+1}-1+1\right)\right)=\frac{1}{N}\left(N-2\sum_{i=1}^N\frac{1}{n_i+1}\right)\leq$$ $$\leq\tfrac{1}{N}\left(N-\tfrac{2\left(\sum\limits_{i=1}^N1\right)^2}{\sum\limits_{i=1}^N(n_i+1)}\right)=1-\tfrac{2N}{\sum\limits_{i=1}^N(n_i+1)}=1-\tfrac{2}{\tfrac{\sum\limits_{i=1}^Nn_i}{N}+1}\leq1-\tfrac{2}{\prod\limits_{i=1}^Nn_i+1}=\tfrac{\prod\limits_{i=1}^Nn_i-1}{\prod\limits_{i=1}^Nn_i+1}.$$