Is the mean of a set of numbers (arranged in decreasing order) always greater its "mean decomposition"?
For simplicity, assume that all the numbers are distinct.
Here's what I mean by "mean decomposition":
Let's say you have three numbers $\alpha, \beta, \gamma$ with $\alpha>\beta>\gamma$. Then the mean of this is $$\mu=\frac{\alpha+\beta+\gamma}3.$$ If we split/decompose this set of numbers as $\{\alpha,\beta\}$ and $\{\gamma\}$ then the mean of the numbers in the first set is $$\mu_1=\frac{\alpha+\beta}2$$ Now define the mean decomposition in this case as $$\nu=\text{mean of}\;\mu_1\,\text{and}\;\gamma=\frac{\frac{\alpha+\beta}2+\gamma}{2}=\frac{\alpha+\beta+2\gamma}{4}$$ Hence the inequality that $$\begin{align}\mu>\nu&\impliedby4(\alpha+\beta+\gamma)>3(\alpha+\beta+2\gamma)\\&\impliedby\alpha+\beta>2\gamma\end{align}$$ is true since $\alpha,\beta>\gamma$.
To generalise this, suppose you have $n$ numbers in descending order $x_1,x_2,\cdots,x_n$.
The mean decomposition would be $$\nu=\text{mean of}(\;\cdots\,(\text{mean of}\;(\text{mean of}\;x_1\,\text{and}\;x_2)\;\text{and}\;x_3)\;\cdots\;\text{and}\;x_n)$$
Where I write $\sum$, it means the sum from $i=1$ to $n$.
Then $$\mu=\frac{\sum x_i}n$$ and it is possible to show that $$\nu=\frac{x_1+\sum2^{i-1}x_i}{2^n}$$
Hence the question can be rephrased as
Is it true that for any integer $n>2$, $$2^n\sum_{i=1}^n x_i>n\left(x_1+\sum_{i=1}^n2^{i-1}x_i\right)\quad?$$
Yes. Your “mean decomposition” is just a weighted average that assigns the elements towards the end of the anti-ordered list heigher weights. $$ \nu = \sum_i w_i\cdot x_i $$ with $$ \sum_i w_i = 1, \qquad w_{i+1} > w_i\ \forall i<n. $$ The ordinary mean can also be interpreted as such a weighted average, but with constant weights: $$ \mu = \sum_i \tfrac1n \cdot x_i $$ Starting from that expression, you can switch to $\nu$ by “redistributing weight to the right”. Which means you're giving more weight to the smaller numbers and less to the higher ones, which obviously reduces the result.