Suppose $f(z)$ is a discrete probability distribution with space $S$. Suppose $g(z),h(z)>0$ for all $z \in S$. Is it true that


$$\prod\_{z \in S}{g(z)^{f(z)}}+\prod\_{z \in S}{h(z)^{f(z)}} \leq \prod\_{z \in S}{[g(z)+h(z)]^{f(z)}}?$$


I'm writing a thesis and this inequality is a lemma to a much broader theorem regarding concavity that I am currently trying to prove. Thanks for your help!



Geometric Expectation Inequality
This is an extension of Mahler's inequality. See <http://en.wikipedia.org/wiki/Mahler%27s_inequality>


We know from the general version of the [AM-GM ineqaulity](http://en.wikipedia.org/wiki/Inequality_of_arithmetic_and_geometric_means) that $\sum f(z) g(z) \geq \Pi g(z)^{f(z)}$ for a discrete distribution $f(z)$ and $g(z) \geq 0$.


In our case, we apply this twice:


1. To $\frac{g(z)}{g(z) + h(z)}$ (Since $g(z), h(z) > 0$, we can do this): $\sum f(z)\frac{g(z)}{g(z) + h(z)} \geq \Pi \left(\frac{g(z)}{g(z)+ h(z)}\right)^{f(z)}$
2. To $\frac{h(z)}{g(z) + h(z)}$ (Since $g(z), h(z) > 0$, we can do this again): $\sum f(z)\frac{h(z)}{g(z) + h(z)} \geq \Pi \left(\frac{h(z)}{g(z)+ h(z)}\right)^{f(z)}$


Adding these two equations gives us:
$$
\sum f(z)\frac{g(z)}{g(z) + h(z)} + \sum f(z)\frac{h(z)}{g(z) + h(z)} \geq \Pi \left(\frac{g(z)}{g(z)+ h(z)}\right)^{f(z)} + \Pi \left(\frac{h(z)}{g(z)+ h(z)}\right)^{f(z)}
$$
Rearranging, we get:
$$
\left(\sum f(z)\frac{g(z) + h(z)}{g(z) + h(z)}\right)\Pi\left(g(z)+ h(z)\right)^{f(z)} \geq \Pi g(z)^{f(z)} + \Pi h(z)^{f(z)}
$$
Since the first term on the left hand side is 1, we get the desired inequality:
$$
\Pi\left(g(z)+ h(z)\right)^{f(z)} \geq \Pi g(z)^{f(z)} + \Pi h(z)^{f(z)}
$$




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Notation shorthand: $\sum$ means $\sum\_{z \in S}$ and $\Pi$ means $\Pi\_{z \in S}$.


