Let $A,D \in \mathbb{R}^{n\times n}$ be two positive definite matrices given by


$$
D =
\begin{bmatrix}
1 & -1 & 0 & 0 & \dots & 0\\
-1 & 2 & -1 & 0 & \dots & 0\\
0 & -1 & 2 & -1 & \dots & 0\\
\vdots & \ddots & \ddots & \ddots & \ddots & 0\\
0 & \dots & 0 & -1 & 2 & -1\\
0 & 0 & \dots & 0 & -1 & 1
\end{bmatrix}, \quad
A =
\begin{bmatrix}
 c\_{1,2} & -c\_{1,2} & 0 & 0 & \dots & 0\\
-c\_{2,1} & c\_{2,1} + c\_{2,3} & -c\_{2,3} & 0 & \dots & 0\\
0 & -c\_{3,2} & c\_{3,2} + c\_{3,4} & -c\_{3,4} & \dots & 0\\
\vdots & \ddots & \ddots & \ddots & \ddots & 0\\
0 & \dots & 0 & -c\_{n-1,n-2} & c\_{n-1,n-2} + c\_{n-1,n} & -c\_{n-1,n}\\
0 & 0 & \dots & 0 & -c\_{n,n-1} & c\_{n,n-1}
\end{bmatrix}
$$
with $c\_{i,j} = c\_{j,i} \in (0,c\_+]$ for all $i,j=1,\dots,n$ for a $c\_+ \in (0,\infty)$.


I would like to prove that independent of the dimension of $n$
$$x^\top A x \le c\_+ x^\top D x$$
holds for all $x\in \mathbb{R}^n$.
If this is not the case does there exist a counter example?


This is somehow related to that the norms of the induced scalar products of the matrices $A$ and $D$ are equivalent with factor $c\_+$.



Prove spectral equivalence of matrices
For $i=1,\dots,n-1$, let $a\_i:=c\_+-c\_{i,i+1}\ge0$. Then, by straightforward calculations with a bit of re-arranging, for $x=(x\_1,\dots,x\_n)\in\mathbb R^n$ we have
$$c\_+ x^\top D x-x^\top A x
=\sum\_{i=1}^{n-1}a\_i(x\_{i+1}-x\_i)^2\ge0.$$


So, your conjectured inequality, $x^\top A x \le c\_+ x^\top D x$, is true.


