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.
Iosif Pinelis
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