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By straightforward calculations with a bit of re-arranging, for $x=(x_1,\dots,x_n)\in\mathbb R^n$ we have $$c_+ x^\top D x-x^\top A x =\sum_{i=1}^{n-1}a_i(x_{i+1}-x_i)^2\ge0,$$ where $a_i:=c_+-c_{i,i+1}\ge0$.

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

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.

For $x=(x_1,\dots,x_n)\in\mathbb R^n$ we have $$c_+ x^\top D x-x^\top A x =\sum_{i=1}^{n-1}a_i(x_{i+1}-x_i)^2\ge0,$$ where $a_i:=c_+-c_{i,i+1}\ge0$.

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

By straightforward calculations with a bit of re-arranging, for $x=(x_1,\dots,x_n)\in\mathbb R^n$ we have $$c_+ x^\top D x-x^\top A x =\sum_{i=1}^{n-1}a_i(x_{i+1}-x_i)^2\ge0,$$ where $a_i:=c_+-c_{i,i+1}\ge0$.

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

For $x=(x_1,\dots,x_n)$ we have $$c_+ x^\top D x-x^\top A x =\sum_{i=1}^{n-1}a_i(x_{i+1}-x_i)^2\ge0,$$ where $a_i:=c_+-c_{i,i+1}\ge0$.

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

For $x=(x_1,\dots,x_n)\in\mathbb R^n$ we have $$c_+ x^\top D x-x^\top A x =\sum_{i=1}^{n-1}a_i(x_{i+1}-x_i)^2\ge0,$$ where $a_i:=c_+-c_{i,i+1}\ge0$.

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