What are the strict local maximisers of $x^T A x$ in the simplex? I am dealing with a matrix in this form.
$A=\begin{pmatrix}
 0&a  & a & a & c & c &c  & c\\ 
 a&  0& a &a  &  c&  c&  c& c\\ 
 a& a &0 &a  &  c& c & c & c\\ 
 a&  a& a &0  & c &c  &c  &c \\ 
 c& c & c & c &  0&  b& b &b \\ 
 c& c & c &  c&  b&0  & b & b\\ 
 c& c & c & c & b & b &0 & b\\ 
 c& c & c & c &  b&  b& b &0 
\end{pmatrix}$
I have $a >b >0.5$ and $c <0.5$.
Want to maximize $x^T A x$, where  $\sum_{i=1}^{8} x_i=1$ and all $x_i \ge 0$  (.i.e, x belongs to the simplex). How to prove that strict local maximizers of the quadratic form  are only $x_1^*=(1/4,1/4,1/4,1/4, 0, 0, 0,0)$ and 
$x_2^*=(0,0,0,0, 1/4, 1/4, 1/4,1/4)$. I can see it intutively as there are 4 blocks in the matrix, but how to prove it.

Consider the gradient of $x^{T}Ax$. Since $A$ is symmetric, the gradient is $2Ax$. Let $x = [x_1\ x_2\ x_3\ldots x_7\ x_8]^{T}$, $l = x_1 + x_2 + x_3 + x_4$ and $r = x_5 + x_6 +x_7 + x_8$, then the gradient is actually (stupid notation..
$$[2(a(l - x_i) + cr)_{i=1, 2, 3, 4} \ 2(cl + b(r - x_i))_{i = 5, 6, 7, 8}]^T$$
Since $a, b > 0$, we can show that any local maximizer $x$ must have $x_1 = x_2 = x_3 = x_4$ and $x_5 = x_6 = x_7 = x_8$. To see this, suppose for example that $x_3 > x_2$. Then $2(a(l - x_3) + cr) < 2(a(l - x_2) + cr)$. This means we should trade $x_3$ for $x_2$ locally (without changing other $x_i$'s) and get larger $x^{T}Ax$. (To show the formally, you can choose to project the gradient to the line where $x_2 + x_3 = 0$ and other dimension being $0$ and see that the gradient is pointing into the simplex.)
Once this is done, the question reduces to deciding the value of $l, r$ as $x_1 = x_2 = x_3 = x_4 = l/4$ and $x_5 = x_6 = x_7 = x_8 = r/4$. The gradient also becomes $[\frac{3}{4}al + cr \ cl + \frac{3}{4}br]^T$. Now discuss cases: there are two end points: $l = 0, r = 1$ and $l = 1, r = 0$, and for other points, we want the gradient to be perpendicular to the $l + r = 1$ line: the two component of the gradient should be equal. 
When $l = 0, r = 1$, the gradient is $[c, \frac{3}{4}b]^T$. For this to be a local maximizer, we want $c \le \frac{3}{4}b$. In other words, $4c - 3b \le 0$. 
When $l = 1, r = 0$, the gradient is $[\frac{3}{4}a, c]^T$. For this to be a local maximizer, we want $c \le \frac{3}{4}a$. In other words, $4c - 3a \le 0$. 
For other points, we want $\frac{3}{4}al + cr = cl + \frac{3}{4}br$. Since $l + r = 1$, we can solve this equation and get:
\begin{align*}
l &= \frac{4c - 3b}{(4c - 3b) + (4c - 3a)} \\
r &= \frac{4c - 3a}{(4c - 3b) + (4c - 3a)}
\end{align*}
We get solutions with $l, r > 0$ exactly when $4c - 3a>0$ and $4c - 3b > 0$.
