A k-cube in $X$ is a function $\psi:\{-1,1\}^k\to (X,d)$. 


An **edge** of a cube is a pair of points $\{\psi(\epsilon\_1),\psi(\epsilon\_2)\}$ in $X$ such that $\epsilon\_1$ and $\epsilon\_2 $ differ in exactly 1 coordinate. Denote the set of all edges by $E$.


An **diagonal** of a cube is a pair of points $\{\psi(\epsilon\_1),\psi(\epsilon\_2)\}$ in $X$ such that $\epsilon\_1=-\epsilon\_2 $ . Denote the set of all diagonal by $D$. 


For any fixed $\epsilon\in\{-1,1\}^k$, denote $E\_{\epsilon}$ the set of edges from $\epsilon$ to $-\epsilon$ obtained by changing the the coordinates sucessively, e.g. $E\_{(1,-1,1)}=\{ \{(1,-1,1),(-1,-1,1)\},\{(-1,-1,1),(-1,1,1)\},\{(-1,1,1),(-1,1,-1)\} \}$. It is clear that $|E\_{\epsilon}|=k$. By triangle inequality and Cauchy-Schwarz we have


$$
k\sum\_{E\_{\epsilon}} \text{edge}^2 \ge \text{diag}^2(\epsilon)
$$


Summing over all $\epsilon$ gives


$$
k\sum\_{E} \text{edge}^2 \ge \sum\_D\text{diag}^2.
$$


Now, assume that there is a (very small) $\theta\in(0,1)$ such that


$$
k\sum\_{E} \text{edge}^2 \ge \sum\_D\text{diag}^2 \ge k(1-\theta)^2\sum\_{E} \text{edge}^2,
$$


we can find $\epsilon\_0$ satisfying


$$
k\sum\_{E\_{\epsilon\_0}} \text{edge}^2 \ge \text{diag}^2(\epsilon\_0) \ge 2^{-k+1}k(1-\theta)^2\sum\_{E} \text{edge}^2.
$$



> 
> It is claimed that, under the assumption, we have
>  $$
> \text{diag}^2(\epsilon\_0)\ge k\left[(1-\theta)^2 \sum\_E\text{edge}^2 - \sum\_{E\backslash E\_{\epsilon\_0}}\text{edge}^2 \right].
> $$
>  Why is it so?
> 
> 
> 


This might be very easy but I really can't think of a way. This inequality is used in the proof of Bourgain-Milman-Wolfson that we can always embed an arbitrary large Hamming cube into a metric space with no non-trivial metric type.


I don't know if this is relevant or not, but $\theta=\frac {\delta^2}{k^22^{k+12}}$ some $\delta\in (0,1)$.



An inequality about embedding of cube into metric spaces
By now surely you already know the answer, but here it is anyway. It follows from the assumption on $\theta$ that
$$
\text{diag}^2(\epsilon\_0)\ge k(1-\theta)^2 \sum\_E\text{edge}^2 - \sum\_{D\_0} \text{diag}^2
$$
where $D\_0$ is the set of diagonals excluding the one determined by $\epsilon\_0$.
We already know that for every $\epsilon$ we have
$$
k\sum\_{E\_{\epsilon}} \text{edge}^2 \ge \text{diag}^2(\epsilon),
$$
so adding up over all the diagonals except the one determined by $\epsilon\_0$, we have
$$
k \sum\_{E\backslash E\_{\epsilon\_0}}\text{edge}^2 \ge \sum\_{D\_0} \text{diag}^2
$$
and the desired inequality follows immediately.


The key here is the remark right after equation (2.1.2) in the Bourgain-Milman-Wolfson paper: "the edges, belonging to identity paths of different diagonals, are all distinct".


