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}[0] & r_{xx}[-1]& \cdots&r_{xx}[-(M-1)] \\r_{xx}[1] &r_{xx}[0] &\cdots&r_{xx}[-(M-2)] \\ \vdots &\vdots & \ddots& \vdots \\ r_{xx}[M-1]&r_{xx}[M-2]&... &r_{xx}[0]\end{array}\right] (2)

Let \mathbf{\alpha}=\left[\begin{array}{ccc} \alpha[0]&\cdots & \alpha[M-1]\end{array}\right]^{T} and \mathbf{x}=\left[\begin{array}{ccc} x[0]&\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\}\geq0
E\left\{\left| \mathbf{\alpha}^{H}\mathbf{x} \right|^{2}\right\}\geq0
E\left\{ \mathbf{\alpha}^{H}\mathbf{x}\mathbf{x}^{H}\mathbf{\alpha} \right\}\geq0

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

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