Let $\Omega = (1- \mu)I_{n\times n} + \mu \boldsymbol{1}_n \boldsymbol{1}_n^\top$, $0\leq \mu<1$. Here $\boldsymbol{1}_n$ denotes the vector of 1's of length $n$. Let $a$ be a positive constant and $D_n=\{x\in \mathbb{R}^n: |x_j|\geq a\; \text{for all $j \in[n]$}\}$. I am trying to obtain tightest possible lower bound for, $$ V_n(a):=\inf_{x\in D_n} x^\top \Omega x .$$ We know that $\lambda_{\min}(\Omega)= (1-\mu)$. also $\Vert x \Vert_2^2\geq na^2$ for all $x\in D$. Hence one obvious lower bound is $V(a)\geq (1-\mu) na^2$. I have shown that this bound is achievable if $n=2k$. For example let $x_j = a .(-1)^j$ for all $j\in[n]$. Then $x^\top \Omega x= (1-\mu) na^2 $. For $n=2k+1$, I am not sure about the tightness of the bound.
This is not a convex optimization problem but I think the symmetry of $\Omega$ should help us in some way. Any help will be appreciated.
Edit: For the case $n= 2k+1$, let $x_j = a.(-1)^j$ for all $j \in [n]$. Then for this choice of $x$ we have $x^\top \Omega x = na^2 \left(1- \frac{n-1}{n+1}\mu \right)$. Hence combining both even and odd case we have $na^2(1-\mu)\leq V_n(a)\leq na^2 \left(1- \frac{n-1}{n+1}\mu \right)$. This suggests that asymptotically the bound is tight, i.e., $$ V_n(a)\sim na^2(1-\mu). $$