Is $\sum_{i=1}^m\prod_{j=1}^nc_j^i\le\prod_{k=1}^n(\sum_{i=1}^m(\frac1{c_k^i}\prod_{j=1}^nc_j^i))^{1/(n-1)}$ true for positive real $c^i_j$?

Question

Context: I am working on a conjectured inequality involving Hausdorff measures, and this a specific case that I believe would help solve the general case.

Fix $n\in\mathbb{N}$. For any $m\in\mathbb{N}$ and for positive real numbers $\{\gamma_j^i\}$, with $j=1,\dots,n$ and $i=1,\dots,m$, is the following true? $$\sum_{i=1}^m \prod_{j=1}^n \gamma_j^i \le \prod_{k=1}^n \left( \sum_{i=1}^m \left(\frac{1}{\gamma_k^i} \prod_{j=1}^n \gamma_j^i \right) \right)^{\frac{1}{n-1}} $$

For clarity note that $\gamma_j^i$ is just another notation for $\gamma_{i,j}.$

I have heuristic reasons to suspect it is in fact true, but wasn't able to prove it. I tried rewriting the inequality as follows:

$$\left( \sum_{i=1}^m \prod_{j=1}^n \gamma_j^i \right)^{n-1} \le \prod_{k=1}^n \sum_{i=1}^m \prod_{\substack{j=1 \\ j\neq k}}^n \gamma_j^i $$ and then analyzing the terms that arise when expanding the power in the l.h.s., but found nothing.

I also tried to use induction in the first formulation of the inequality. The case $m=1$ is trivial, but then it comes down to

\begin{align*} \sum_{i=1}^m \prod_{j=1}^n \gamma_j^i & = \sum_{i=1}^{m-1} \prod_{j=1}^n \gamma_j^i + \prod_{j=1}^n\gamma_j^m \\ & \le \prod_{k=1}^n \left( \sum_{i=1}^{m-1} \left(\frac{1}{\gamma_k^i} \prod_{j=1}^n \gamma_j^i \right) \right)^{\frac{1}{n-1}} + \prod_{j=1}^n\gamma_j^m \\ \end{align*} which got me stuck.

Another approach: raising both sides of the original inequality to $\frac1n$ results in a geometric mean on the r.h.s.. However, using the HM-GM inequality is not the correct path, since the inequality is not true when replacing the geometric mean with the harmonic one.

Finally, notice that, at least in the case where $\gamma_j^i=\gamma$ for each $i$ and $j$, the inequality reduces to $1\le m^{\frac{1}{n-1}}$, which is true since $m\ge 1$. Another solved case is when $m=2$ and $n=3$, which can be found here. In this specific case, the inequality is proven to be strict, and I believe it to always be this way, unless $m=1$.

EDIT: As @Noctis pointed out in the comments, it is not restrictive to assume that $\gamma_j^i\in(0,1)$ for every $i$ and $j$, reducing the space of possible combinations.

Any help would be greatly appreciated, thank you.

Answer 1

Problem : $$\text{LHS}=\sum_{i=1}^m \prod_{j=1}^n \gamma_j^i \le \prod_{k=1}^n \left( \sum_{i=1}^m \left(\frac{1}{\gamma_k^i} \prod_{j=1}^n \gamma_j^i \right) \right)^{\frac{1}{n-1}}=\text{RHS}$$ Let $$\beta^i_k=\left(\frac{1}{\gamma_k^i} \prod_{j=1}^n \gamma_j^i\right)^\frac{1}{n-1}\implies \prod_{j=1}^n \gamma_j^i=\prod_{j=1}^n \beta_j^i \tag{1}$$ We have $$\text{RHS}=\prod_{k=1}^n \left( \sum_{i=1}^m (\beta_k^i)^{n-1} \right)^{\frac{1}{n-1}}=\prod_{k=1}^n\|\beta_k\|_{n-1}\ge\|\beta_1\cdot\cdot\cdot\beta_n\|_{1-\frac{1}{n}}$$ Note that by Generalized Hölder's inequality:
For $q_1,..,q_n,r>0$ where $\sum_{k=1}^n\frac{1}{q_n}=\frac{1}{r}$ $$\|f_1\cdot\cdot\cdot f_n\|_r\le\|f_1\|_{q_1}\cdot\cdot\cdot\|f_n\|_{q_n}$$ So let $\forall k,q_k=n-1$ implies we need to set $r=1-\frac{1}{n}$ for the inequality to hold true. Now substituting $(1)$ into $\text{LHS}$, we have $$\text{LHS}=\|\beta_1\cdot\cdot\cdot\beta_n\|_1\le\|\beta_1\cdot\cdot\cdot\beta_n\|_{1-\frac{1}{n}}\le\text{RHS} \ \ \square$$ Clarification:
$\beta_j^i,\gamma_j^i$ are numbers indexed with $i,j$
$\beta_k:=(\beta_k^1,..,\beta_k^m);\beta_1\cdot\cdot\cdot\beta_k:=(\beta_1^1\cdot\cdot\cdot\beta_k^1,..,\beta_1^m\cdot\cdot\cdot\beta_k^m)$

References
[1] Hölder's inequality for multiple functions
[2] Monotonicity of $\|x\|_p$