Showing $u^T M u \geq v^TMv$ when $M$ is symmetric PD and $u,v$ are $0-1$ vectors Let $M$ be a $n \times n$ symmetric positive definite matrix. Let $u$ and $v$ be vectors of length $n$ with entries consisting $n-m_u$ (or $n-m_v$) $0$'s and $m_u $ (or $m_v$) $1$'s, where $m_u,m_v \in \{1,...,n-1\}$. Sort $u$ so that the first $n-m_u$ entires of $u$ are $0$'s and the last $m_u$ entries are $1$'s. Sort $v$ in the same way. Suppose $m_u>m_v$. Is the following weak inequality true? 
(As shown below by @Niki Di Giano, this is not true)
$$ u^T M u \geq v^TMv $$ 
 A: Consider : $$
M=
  \begin{bmatrix}
    2 & -1 & 0 \\
    -1 & 2 & -1 \\
    0 & -1 & 2
  \end{bmatrix}, u  = \begin{bmatrix}
    1 & 1 & 1 \end{bmatrix} ^T,   v  = \begin{bmatrix}
    0 & 1 & 1 \end{bmatrix} ^T                       
$$
M is positive definite and : 
$$ u^T M u = v^T M v = 2 $$
so it's not a strict inequality (is $m_u \lt n$ ?). I couldn't find an example where the inequality is not working though.
A: Another example where the inequality fails:
The matrix $\begin{bmatrix}3&-2\\-2&3\end{bmatrix}$ is positive definite. Also, $\begin{bmatrix}1\\1\end{bmatrix}>\begin{bmatrix}1\\0\end{bmatrix}$. But $$\begin{bmatrix}1&1\end{bmatrix}\begin{bmatrix}3&-2\\-2&3\end{bmatrix}\begin{bmatrix}1\\1\end{bmatrix}=2<3=\begin{bmatrix}1&0\end{bmatrix}\begin{bmatrix}3&-2\\-2&3\end{bmatrix}\begin{bmatrix}1\\0\end{bmatrix}$$ which violates the inequality.
A: The matrix $M$:
$$M =
\begin{bmatrix}
1 & 0 & 0\\
0 & 10 & - 9\\
0 & -9 & 10\\
\end{bmatrix}
$$
Yields $u^T M u = 2$ and $v^T M v = 10$ for the vectors $u = (0, 1, 1)$ and $v =(0, 0, 1)$ respectively.
