In [1, p. 62 exercise 3.14] it is requested to proof prove that the autocorrelation matrix given by [1, eq. 3.46] is also positive semidefinite. This shall be done by usage of the definition of the semidefinite property of the ACF in [1, eq. 3.45].
Solution: A matrix $\mathbf{A}$ is positive semidefinite if $\mathbf{x}^{H}\mathbf{A}\mathbf{x}\geq 0$. And the autocorrelation matrix given by ([1, eq. 3.46]) is: $\mathbf{R}_{xx}$ $=$ $E\left\{\mathbf{x}\mathbf{x}^{H}\right\}$ (1) $=$ $\left[\begin{array}{cccc}r_{xx} & r_{xx}[-1]& \cdots&r_{xx}[-(M-1)] \\r_{xx} &r_{xx} &\cdots&r_{xx}[-(M-2)] \\ \vdots &\vdots & \ddots& \vdots \\ r_{xx}[M-1]&r_{xx}[M-2]&... &r_{xx}\end{array}\right]$ (2)

Let $\mathbf{\alpha}=\left[\begin{array}{ccc} \alpha&\cdots & \alpha[M-1]\end{array}\right]^{T}$ and $\mathbf{x}=\left[\begin{array}{ccc} x&\cdots & x[M-1]\end{array}\right]^{T}$ then from [1, eq. 3.45] we obtain: $E\left\{\left| \sum\limits_{n=0}^{M-1}\alpha^{\ast}[n]x[n] \right|^{2}\right\}$ $\geq$ $0$ $E\left\{\left| \mathbf{\alpha}^{H}\mathbf{x} \right|^{2}\right\}$ $\geq$ $0$ $E\left\{ \mathbf{\alpha}^{H}\mathbf{x}\mathbf{x}^{H}\mathbf{\alpha} \right\}$ $\geq$ $0$ $\mathbf{\alpha}^{H}E\left\{\mathbf{x}\mathbf{x}^{H}\right\}\mathbf{\alpha}$ $\geq$ $0$ $\mathbf{\alpha}^{H}\mathbf{R}_{xx}\mathbf{\alpha}$ $\geq$ $0$

Thus $\mathbf{R}_{xx}$ is positive semidefinite. QED.

 Steven M. Kay: “Modern Spectral Estimation – Theory and Applications”, Prentice Hall, ISBN: 0-13-598582-X.