Let $C^n=\{0,1\}^n$ be a metric space (Hamming Cube). The distance on $C^n$ is defined by $$ d(\varepsilon,\varepsilon'):=|\{j:\varepsilon_j\ne\varepsilon'_j\}|, $$ $\varepsilon=(\varepsilon_1,\dots,\varepsilon_n)$.

Let $s,k$ be integers such that $sk=n$. We divide each $\varepsilon$ into $k$ blocks, each of lenght $s$ i.e. $$\underbrace{ \underbrace{\varepsilon_1, \varepsilon_2,\dots,\varepsilon_s}_{s\text{ times}}, \underbrace{\varepsilon_{s+1}, \varepsilon_{s+2},\dots,\varepsilon_{2s}}_{s\text{ times}},\dots, \underbrace{\varepsilon_{(k-1)s+1}, \varepsilon_{(k-1)s+2},\dots,\varepsilon_{n}}_{s\text{ times}} }_{k\text{ times}} $$ Denote the $i^{\text{th}}$ block by $I_i:=\{(i-1)s+1,(i-1)s+2,\dots,is \}$.

Define $\varepsilon_{I_i}$ by

$$
(\varepsilon_{I_i})_j:=
\begin{cases}1-\varepsilon_j &;\ j\in I_i \\
\varepsilon_j &;\ j\ne I_i
\end{cases}
$$
to be the *swap* of the $i^{\text{th}}$ block of $\varepsilon$. For example let $n=6,k=3,s=2$, then
$$\begin{align}
(000000)_{I_1}&=(110000)\\
(000000)_{I_2}&=(001100)\\
(011001)_{I_1}&=(101001).
\end{align}$$

Any permutation $\sigma\in S_n$ induces a transformation on $\varepsilon$ in an obvious way: $$ (\sigma\varepsilon)_j := \varepsilon_{\sigma(j)}. $$

Let $f:C^n \to X$, where $X$ is another metric space. Denote the

diagonal map$\delta$ defined by $$ \delta(\varepsilon):=d(f(\varepsilon),f(1-\varepsilon)) $$ and a another map $\phi_i(\varepsilon,\sigma)$ defined by $$ \phi_i(\varepsilon,\sigma):=d(f(\sigma\varepsilon),f(\sigma\varepsilon_{I_i})). $$

There are two claims $$\begin{align} \sum_{\sigma,\varepsilon}\phi_i(\varepsilon,\sigma) &= \sum_{\sigma,\varepsilon}\phi_j(\varepsilon,\sigma)\quad\text{for all}\ i,j\quad\text{and} \\ \sum_{\varepsilon}\delta(\varepsilon) &\le \frac 1{n!}\sum_{i=1}^k \sum_{\sigma,\varepsilon}\phi_i(\varepsilon,\sigma). \end{align}$$

The first one is not so hard, but I cannot wrap my head around the second one yet. I believe a prove would go along the line of using triangle inequality repeatedly, but right now I am confused by $\phi_i(\varepsilon,\sigma)$.

Is there a nice way to look at $\phi_i(\varepsilon,\sigma)$? What is it intuitively?

**EDIT**: To generalize the inequality, we can write
$$
E_1 := \frac 1{n!}\sum_{\sigma,\varepsilon}\phi_i(\varepsilon,\sigma)
$$
and, since $E_1$ does not depend on the choice of $i$, the previous inequality become
$$
\sum_{\varepsilon}\delta(\varepsilon) \le k E_1 = \frac ns E_1.
$$
Now define $\phi_{i,j}:=d(f(\sigma\varepsilon),f(\sigma\varepsilon_{I_i\cup I_j}))$, where $\varepsilon_{I_i\cup I_j}$ is the swapping of both $i^{\text{th}}$ and $j^{\text{th}}$ block,
$$
E_2 := \frac 1{n!}\sum_{\sigma,\varepsilon}\phi_{i,j}(\varepsilon,\sigma)
$$
for $i\ne j$. How do we prove
$$
\sum_{\varepsilon}\delta(\varepsilon) \le \frac n{2s} E_2
$$
and, even more generally,
$$
\sum_{\varepsilon}\delta(\varepsilon) \le \frac n{ts} E_t
$$
where $E_t$, $1\le t\le k$, is defined using $\varepsilon_{I_{i_1}\cup I_{i_2}\cup\cdots\cup I_{i_t}}$ with $i_1<i_2<\dots<i_t$?

This is not hard when $ts|n$ but for general $t$ I cannot see it.