Intuition for inequality involving permutation and Hamming Cube 
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.
 A: If we took $\phi_i(\varepsilon, (1))$, where $(1)$ is the identity permutation, we'd have $d(f(\varepsilon), f(\varepsilon_{I_i}))$: the distance we go when we change the $i$-th block of $\varepsilon$. 
But there's nothing special about the order of the coordinates of $\varepsilon$, nor about the way we partition them into blocks. So we introduce the permutation $\sigma$ to average out over all of these. In fact, if we consider the values of $\phi_i(\sigma^{-1}\varepsilon, \sigma)$ as $\sigma$ ranges over all of $S_n$, we'd see every possible way to change $s$ coordinates of $\varepsilon$ a total of $n!/\binom{n}{s}$ times.
But there's a reason that we keep blocks around, rather than just permuting over sets of $s$ coordinates. We can use blocks to define the sequence $\varepsilon^{(0)}, \varepsilon^{(1)}, \dots, \varepsilon^{(k)}$, where:


*

*we start with an $\varepsilon^{(0)} = \varepsilon$,

*then flip the blocks one at a time by taking $\varepsilon^{(i)} = \varepsilon^{(i-1)}_{I_i}$,

*and as a result end up at $\varepsilon^{(k)} = 1 - \varepsilon$.


This is a particular choice of $k$-step path from $\varepsilon$ to $1 - \varepsilon$, and $$\phi_1(\varepsilon^{(0)}, (1)) + \phi_2(\varepsilon^{(1)}, (1)) + \dots + \phi_k(\varepsilon^{(k-1)}, (1))$$ will tell us the length of that path in the embedding $C^n \to X$. If we now let $\varepsilon$ and $\sigma$ vary, then
$$\phi_1(\varepsilon^{(0)}, \sigma) + \phi_2(\varepsilon^{(1)}, \sigma) + \dots + \phi_k(\varepsilon^{(k-1)}, \sigma)$$
will vary over the lengths of all possible $k$-step paths, where we can start at any $\varepsilon$ and flip the coordinates in any possible groups of $s$ at a time, until we end up at $1-\varepsilon$.
By the triangle inequality, $$\delta(\varepsilon) \le \phi_1(\varepsilon^{(0)}, \sigma) +  \dots + \phi_k(\varepsilon^{(k-1)}, \sigma)$$ since the $k$-step path can be no shorter than a direct path from $\varepsilon$ to $1-\varepsilon$. So if we average over all the paths, then we must have
$$\frac1{2^n} \sum_{\varepsilon} \delta(\varepsilon) \le \frac1{2^n\,n!}\sum_{\sigma,\varepsilon} \left(\phi_1(\varepsilon^{(0)}, \sigma) +  \dots + \phi_k(\varepsilon^{(k-1)}, \sigma)\right).$$ The rest of the work is just rearranging the sum on the right-hand side to "forget" about the existence of $k$-step paths.
