Prove spectral equivalence of matrices 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 $c_+$.

This seems a counterexample: $c_+=1$ (which you can assume without loss of generality), $n=2$, $A = \begin{bmatrix}2 & -\varepsilon \\\ -\varepsilon & 2\end{bmatrix}$, $x = \begin{bmatrix}1 \\ 1\end{bmatrix}$ gives $x^*Dx=2, x^*Ax = 4 - 2\varepsilon$, so the inequality is reversed.
This is not just a wrong sign, since $x = \begin{bmatrix}1 \\ -1\end{bmatrix}$ gives an inequality with the opposite sign. There just does not seem to be an inequality of that kind valid for every $x$.
