How to show that two similar expressions involving natural logs of summations of exponents are not the same?

Question

I've come across in various papers two slightly different ways to calculate supposedly the same physical quantity. Now, numerically in Excel it's easy to show that for real-world input values the results almost always are different, but I need to show rigorously/analytically that in non-trivial cases they are indeed not equivalent, or alternatively give the unique non-trivial conditions when they actually do agree.

I thought the solution might involve the "LogSumExp function" I came across on Wikipedia or the expectation (mean) of some modified exponential probability distribution but it's a beyond my abilities.

Specifically, how can you demonstrate rigorously/analytically that (presumably??) in all non-trivial cases the two similar expressions $(1)$ and $(2)$ below involving natural logs of summations of exponents yield different results for a given finite (integer) number $n$ of $x_i$ values where $x_i \in \mathbb{R}: x_i > 0$ (NB: it's possible for any two or more $x_i$ values to be equal)? $$\frac{-1}{\ln(\frac{1}{n}{\sum_{i=1}^n e^\frac{-1}{x_i}})} \tag{1}$$ $$\frac{-1000}{\ln(\frac{1}{n}{\sum_{i=1}^n e^\frac{-1000}{x_i}})} \tag{2}$$

A trivial case is where all the $x_i$ values are equal, and the two expressions will yield the same result. But in general I don't think this holds, but I can't prove it.

I hope this makes sense – I'm only a chemist not a mathematician, statistician or physicist!

Answer 1

A little algebraic manipulation will simplify the task. Write $t_i:=e^{-1/x_i}$. Then by hypothesis each $t_i$ is between $0$ and $1$. If you divide both (1) and (2) by $1000$ and use the property $a\ln u=\ln u^a $, you find that you are comparing $\left(\frac1n\sum t_i\right)^a$ to $\frac1n\sum t_i^a$, where $a:=1000$, or equivalently you are comparing $$ f\left(\frac1n\sum t_i\right)\tag{1a}$$ to $$\frac1n\sum f(t_i),\tag{2a}$$ where the function $f$ is defined as $$f(t):=t^a.$$ Since the second derivative of $f$ is strictly positive for $t$ between $0$ and $1$, the function $f(t)$ is strictly convex for $t$ between $0$ and $1$. One consequence of strict convexity:

If $f$ is strictly convex, then $$f\left(\frac1n\sum t_i\right)\le\frac1n\sum f(t_i),$$ with equality if and only if all the $t_i$ values are equal.

In other words, (1a) is strictly less than (2a) except when all the $t$'s are equal. Translating back to your context, this means (if I've done my algebra correctly) that (1) is strictly less than (2) except when all the $x_i$ values are equal.