Prove this inequality with Cauchy-Schwarz inequality Let $x_{1},x_{2},\cdots,x_{n}>0$, show that
$$\left(\sum_{k=1}^{n}x_{k}\cos{k}\right)^2+\left(\sum_{k=1}^{n}x_{k}\sin{k}\right)^2\le \left(2+\dfrac{n}{4}\right)\sum_{k=1}^{n}x^2_{k}$$
I can prove it when $2+\dfrac{n}{4}$  takes the place of $n$,
It seems we should use Cauchy-Schwarz inequality: 
$$\left(\sum_{k=1}^{n}x_{k}\cos{k}\right)^2\le\sum_{k=1}^{n}x^2_{k}\sum_{k=1}^{n}\cos^2{k}\tag{1}$$
$$\left(\sum_{k=1}^{n}x_{k}\sin{k}\right)^2\le\sum_{k=1}^{n}x^2_{k}\sum_{k=1}^{n}\sin^2{k}\tag{2}$$
Adding $(1),(2)$ we have:
$$\left(\sum_{k=1}^{n}x_{k}\cos{k}\right)^2+\left(\sum_{k=1}^{n}x_{k}\sin{k}\right)^2\le\sum_{k=1}^{n}x^2_{k}\sum_{k=1}^{n}(\cos^2{k}+\sin^2{k})=n\sum_{k=1}^{n}x^2_{k}$$
 A: Under what circumstances it is verified that the eigenvalues ​​of 
$$
Q_n = a_n\sum_{k=1}^n x_k^2 -\left(\sum_{k=1}^n x_k\sin k\right)^2-\left(\sum_{k=1}^n x_k\cos k\right)^2
$$
are all non-negative?
$$
Q_n = X^{\dagger}M_n X
$$
where 
$$
M_n = \left(
\begin{array}{ccccc}
       a_n-1 & -\cos (1) & \cdots & -\cos (n-2) & -\cos (n-1) \\
 -\cos (1) &       a_n-1 & \cdots & -\cos (n-3) & -\cos (n-2) \\
 \vdots & \vdots & \vdots & \vdots & \vdots \\
 -\cos (n-2) & -\cos (n-3) & \cdots & a_n-1       & -\cos (1) \\
 -\cos (n-1) & -\cos (n-2) & \cdots & -\cos (1) & a_n-1 \\
\end{array}
\right)
$$
The $M_n$ eigenvalues are
$$
\{a_1-1\}\\
\{a_2-1-\cos (1),a_2-1+\cos (1)\}\\
\{a_3,a_3-2-\cos (2),a_3-1+\cos (2)\}\\
\{a_4,a_4,a_4-2-\cos (3)-\cos (1),a_4-2+\cos (3)+\cos (1)\}\\
\{a_5,a_5,a_5,a_5-3-\cos (4)-\cos(2),a_5-2+\cos (4)+\cos (2)\}\\
\{a_6,a_6,a_6,a_6,a_6-3-\cos (5)-\cos (3)-\cos (1),a_6-3+\cos (5)+\cos (3)+\cos (1)\}\\
\{a_7,a_7,a_7,a_7,a_7,a_7-4-\cos (6)-\cos (4)-\cos(2),a_7-3+\cos (6)+\cos (4)+\cos (2)\}\\
\{a_8,a_8,a_8,a_8,a_8,a_8,a_8-4-\cos (7)-\cos (5)-\cos (3)-\cos (1),a_8-4+\cos (7)+\cos(5)+\cos (3)+\cos(1)\}\\
\{a_9,a_9,a_9,a_9,a_9,a_9,a_9,a_9-5-\cos (8)-\cos (6)-\cos (4)-\cos (2),a_9-4+\cos (8)+\cos(6)+\cos (4)+\cos (2)\}\\
\vdots
$$
and the conditions on $a_n$ such that all the eigenvalues are positive 
$$
a_1 > 1\\
a_2 > 1+\cos(1)\\
a_3 > 1+\cos(2)\\
a_4 > 2-\cos (1)-\cos (3)\\
a_5 > 2-\cos (2)-\cos (4)\\
a_6 > 3-\cos (1)-\cos (3)-\cos (5)\\
a_7 > 4+\cos (2)+\cos (4)+\cos (6)\\
a_8 > 4+\cos (1)+\cos (3)+\cos (5)+\cos (7)\\
a_9 > 5+\cos (2)+\cos (4)+\cos (6)+\cos (8)\\
\vdots
$$
Follows a plot showing in blue $\{a_k\}$ and in red $\{2+\frac k4\}$

As we can observe, for $k \le 6$ the factor $\color{red}{(2+\frac k4)}$ is well placed but for $k \ge 7$ remains a strong doubt. In green a factor which agrees with the eigenvalues positiveness. $\color{green}{(\frac 23+\frac k2)}$
