@vidyarthi : imgur.com/a/XuyKFkD

@vidyarthi no need for positive. You need to work a bit with pen and paper. Hint : $\sum_{i=1}^3\sum_{j=1}^3 c_ic_j = c_1^2 + c_2^2 + c_3^2 + 2c_1c_2 + 2c_2c_3 + 2c_3c_1 = (c_1 + c_2 + c_3)^2$

also write $2\cos{\theta} = e^{2\pi i\theta} + e^{-2\pi i\theta}$

@vidyarthi : any $c = [c_1,c_2,...c_n] \in \mathbb{R}^n \setminus \{0\}^n$. I will expand the summation later when I get time. Its just an interchange of order of summations and an algebraic manipulation.

Thanks for your valuable comments and answer.

Your counter example seems valid to me. Luckily in my problem $C$ goes as $\frac{1}{\lambda}$...perhaps $O$ is not suitable here, I should have used $\omega$ or something. I have a closed form expression for $C$ which is $\frac{1}{\lambda}$. Thank you for the answer.

$M$ is positive definite.

Your example still assumes $\rho(M) = 0$ which is not possible.

In question I had stated $A+C$ is known to be $psd$ so $M$ is a pd. Hope that solves the problem.