I'm a bit stuck on a proof that seems to hold when I test in in graphing calculators, would love to hear some tips on strategies I might use.
Suppose we have two finite sets $\Omega$ and $\Phi$, with generic elements $\omega$ and $\phi$. We have four indexed sequences $(L_\omega)\in(0,1)^{\Omega}$, $(Z_{\phi})\in(0,1)^{\Phi}$, $(Q_{\phi})\in\mathbb{R}_{>0}^{\Phi}$, $(x_{\phi,\omega})\in(0,1)^{\Phi\times\Omega}$ such that \begin{align*} &\sum_{\omega\in\Omega}L_\omega=1\\ % &\sum_{\phi\in\Phi} Z_\phi=1\\ % &\forall \omega\in\Omega\left(\sum_{\phi\in\Phi}x_{\phi,\omega}=1\right)\\ % &\forall \phi\in\Phi\left(\sum_{\omega\in\Omega}L_{\omega}x_{\phi,\omega}=Z_{\phi}\right)\\ \end{align*} then I need to prove that \begin{align} \sum_{\omega\in\Omega}L_{\omega}\left[\sum_{\phi\in\Phi}\left({\frac{x_{\phi,\omega}}{1+\frac{Q_\phi}{x_{\phi,\omega}}}-\frac{x_{\phi,\omega}}{1+\frac{Q_\phi}{Z_\phi}}}\right)\right] \left[\sum_{\phi\in\Phi}\left({\frac{x_{\phi,\omega}}{1+\frac{Q_\phi}{x_{\phi,\omega}}}-\frac{Z_\phi}{1+\frac{Q_\phi}{Z_\phi}}}\right)\right]>0\tag{1}\label{eq: together} \end{align} excluding the cases in there exist two states $\omega,\omega'$ in which $x_{\omega,\phi}=x_{\omega',\phi}$ for all $\phi$ or there exists some state $\omega$ such that $x_{\omega,\phi}=Z_{\phi}$ for all $\phi$.
The main problem is that while I know \begin{align} &\sum_{\omega\in\Omega}L_{\omega}\left[\sum_{\phi\in\Phi}\left({\frac{x_{\phi,\omega}}{1+\frac{Q_\phi}{x_{\phi,\omega}}}-\frac{x_{\phi,\omega}}{1+\frac{Q_\phi}{Z_\phi}}}\right)\right] >0\\ \end{align} and \begin{align} &\sum_{\omega\in\Omega}L_{\omega}\left[\sum_{\phi\in\Phi}\left({\frac{x_{\phi,\omega}}{1+\frac{Q_\phi}{x_{\phi,\omega}}}-\frac{Z_{\phi}}{1+\frac{Q_\phi}{Z_\phi}}}\right)\right]>0 \end{align} by Jensen's inequality, for a given $\omega$ these might be negative or positive, and so I'm having trouble saying anything about the convexities or derivatives of the product-sum I'm trying to prove without know these signs for each $\omega$. And in fact, ultimately what I'm trying to prove is the following:
\begin{align} \exists\omega\in\Omega\left(\left(\sum_{\phi\in\Phi}\left({\frac{x_{\phi,\omega}}{1+\frac{Q_\phi}{x_{\phi,\omega}}}-\frac{x_{\phi,\omega}}{1+\frac{Q_\phi}{Z_\phi}}}\right)>0\right)\land \left(\sum_{\phi\in\Phi}\left({\frac{x_{\phi,\omega}}{1+\frac{Q_\phi}{x_{\phi,\omega}}}-\frac{Z_\phi}{1+\frac{Q_\phi}{Z_\phi}}}\right)>0\right)\right).\tag{2} \end{align} But in the graphing calculator, I was not always able to identify a pattern that identified which $\omega$ this would be (in particular, it was not necessarily always the one with the maximum $\sum_{\phi\in\Phi}\frac{x_{\phi,\omega}}{1+\frac{Q_\phi}{x_{\phi,\omega}}}$, which would be a natural conjecture). So I found out that the sum in expression (1) was always positive in my graphing software and with a little work, I could prove it was a sufficient condition for (2). Thus I could prove (2) without constructively identifying the $\omega$ that it holds for by proving (1).
Clearly it is exactly zero in the case that I ruled out where all $x_{\phi,\omega}=Z_{\phi}$, so if I could prove this is a unique minimum that would work. Also, I noticed that if you stuck a positive scalar onto $Q$, i.e. define a $\tilde{Q}_{\phi}$ such that $Q_{\phi}=k\tilde{Q}_{\phi}$ for $k>0$, then (1) looks quasi-concave in $k$, where obviously it approaches $0$ as $k\to0$ or $k\to\infty$, so proving this quasi-concavity would also be sufficient.
Even with just sets of size three $|\Omega|=|\Phi|=3$, the expression ends up with 51 terms each with unique denominators, so I am having trouble seeing a clever way to collect terms and get a Jensen's Inequality expression. Anyone know any other proof techniques or theorems that might be helpful?