Show that $[\prod_{i=1}^N n_i] + 1 \geq \sqrt{N}(\bar{n} + 1)$

Question

Following on from my question here, I am trying to show now that $\left[\prod_{i=1}^N n_i\right] + 1 \geq \sqrt{N}(\bar{n} + 1)$, where $\bar{n} = \frac{1}{N}\sum_{i=1}^N n_i$ and $n_i \in \mathbb{N}, n_i \geq 2$. I tried to show it by induction but I was not having much luck. I am not mathematician but it seems like it should be reasonably easy to prove, but alas! I have confirmed with some simulations (simulating $n_i$ for various $N$) with $N$ up to around 20 before overflow errors, so it looks promising that it should be true, but I am having a really difficult time proving it.

Answer 1

We'll prove this by induction on $N\geq 2$ (case $N=1$ is trivial). Let $x=n_N$ and $$f(x) = \left(\prod_{k=1}^{N-1} n_k - \dfrac{\sqrt{N}}{N}\right)x + 1 - \sqrt{N}\left(\dfrac{\sum_{k=1}^{N-1}n_k}{N} + 1\right)$$ We want to prove that $f([2,+\infty[) \in \mathbb{R}_+$ (initial inequality written differently).

Notice that $$ f'(x) = \prod_{k=1}^{N-1} n_k - \dfrac{\sqrt{N}}{N} \geq 2^{N-1} - \dfrac{\sqrt{N}}{N} \geq 0$$ Hence $f$ is increasing ie. $f(x) \geq f(2)$. Finally, \begin{align*} f(2) &= 2\left(\prod_{k=1}^{N-1} n_k - \dfrac{\sqrt{N}}{N}\right) + 1 - \sqrt{N}\left(\dfrac{\sum_{k=1}^{N-1}n_k}{N} + 1\right)\\ &= 2\prod_{k=1}^{N-1} n_k + 2 - 1 - \sqrt{N}\left(\dfrac{\sum_{k=1}^{N-1}n_k}{N} + \dfrac{2}{N} +1\right)\\ &\overset{IH}\geq 2\sqrt{N-1}\left(\dfrac{\sum_{k=1}^{N-1}n_k}{N-1} + 1\right) - \sqrt{N}\left(\dfrac{\sum_{k=1}^{N-1}n_k}{N} + \dfrac{2}{N} +1\right)-1\\ &= \left(\sum_{k=1}^{N-1}n_k\right)\left(2\dfrac{\sqrt{N-1}}{N-1}-\dfrac{\sqrt{N}}{N}\right)+2\sqrt{N-1}-1-\dfrac{2\sqrt{N}}{N}-\sqrt{N}\\ &\geq 2(N-1)\left(2\dfrac{\sqrt{N-1}}{N-1}-\dfrac{\sqrt{N}}{N}\right)+2\sqrt{N-1}-1-\dfrac{2\sqrt{N}}{N}-\sqrt{N}\\ &= 6\sqrt{N-1}-1-\sqrt{N}-2\dfrac{\sqrt{N}}{N}\left(1+(N-1)\right)\\ &= 6\sqrt{N-1} - 3\sqrt{N} - 1 \geq 0 \end{align*} The last inequality holds for $N\geq 2$.

There's probably an easier solution.

Answer 2

Set $n_i = x_i + 2$ with $x_i \ge 0$. Then $$ 1+\prod_{i=1}^N n_i = 1+\prod_{i=1}^N (x_i+2) \ge 1+2^N + 2^{N-1}\sum_{i=1}^N x_i \, . $$ Now $1+2^N \ge 3 \sqrt N$ (proof below) and $2^{N-1} \ge 1 \ge \frac{1}{\sqrt N}$, therefore the expression is $$ \ge 3 \sqrt N + \frac{1}{\sqrt N}\sum_{i=1}^N x_i = \sqrt N \left(3 + \frac 1N \sum_{i=1}^N x_i \right) = \sqrt N \left(1 + \frac 1N \sum_{i=1}^N n_i \right) \, . $$

---

It remains to show that $1+2^N \ge 3 \sqrt N$ for $N \ge 1$. For $N=1, 2, 3$ this can be verified directly. If $N \ge 4$ then $N=u^2$ with $u \ge 2$ and, using Bernoulli's inequality, $$ 1+2^N - 3 \sqrt N = 1+(1+1)^{u^2} - 3u \ge 1+ (1+u^2) - 3u = (u-2)(u-1) \ge 0 \, . $$