The probability that a point is close to a hyperplane that is orthogonal to the principal diagonal must be very big. Let $\mu^n$ be the uniform probability measure on the $n$-dimensional cube $[-1,1]^n$. Let $H \in\mathbb{ R}^n$ be the hyperplane orthogonal to the principal diagonal, i.e., $H = (1,\cdots ,1)^\perp$. For any $r > 0$, we further define
$$A_{H,r}:=\{x\in[-1,1]^n,\text{dist}(x,H)\le r\},$$
where $\text{dist}(x,H)$ represents the distance from the point $x$ to the hyperplane $H$. Show that for any constant $\epsilon > 0$, the following estimate hold for all sufficiently large $n$
$$\mu^n(A_{H,n^\epsilon})\ge1-e^{-n^{\epsilon/2}}$$
$$$$
It is not hard to prove the inequality $\mu^n(A_{H,n^\epsilon})\ge1-n^{-2\epsilon}$. However the stronger inequality form this problem is too strong with $e$ as exponential. I think to prove the stronger inequality we may need further make use of the "hyperplane orthogonal to the principal diagonal" hypothesis. Are there any ways of deriving this stronger inequality?
 A: This follows from classical concentration inequalities (used for large deviation estimates), in particular the Chernoff bound. However, I think it is more instructive to illustrate the technique used to prove the Chernoff bound by deriving the desired inequality from scratch, as follows.
Let $X_1,\ldots,X_n$ be independent random variables uniformly distributed on $[-1,1]$. Then
$$
\mathbb P(|X_1+\cdots+X_n|>t)\leq 2\mathbb P(X_1+\cdots+X_n>t).
$$
By Markov's inequality,
$$
\mathbb P(e^{aX_1+\cdots+aX_n}>e^{at})\leq \mathbb E[e^{aX_1}]^n e^{-at}=\sinh(a)^na^{-n}e^{-at}.
$$
From the Taylor expansion of $\sinh(a)$ it follows that for all $|a|$ less than some absolute constant, the inequality $\sinh(a)\leq a+a^3$ holds, thus for all such $a$
$$
\mathbb P(X_1+\cdots+X_n>t)\leq (1+a^2)^ne^{-at}\leq e^{na^2-at}.
$$
The quadratic in the exponent is minimized when $a=t/2n$. Choose $n$ sufficiently large ensures that $a$ is small enough for the previous inequality to apply, thus for all such $n$
$$
\mathbb P(X_1+\cdots+X_n>t)\leq e^{-t^2/4n}.
$$
Therefore
$$
\mu^n(A_{H,n^{\epsilon}})=\mathbb P\left(|X_1+\cdots+X_n|\leq n^{\epsilon+1/2}\right)\geq 1-2e^{-n^{2\epsilon}/4}.
$$
Since $\epsilon>0$, we have that $2\epsilon>\epsilon/2$ and thus the inequality is stronger than the one desired in the question.
