Exercise [1, p. 14 exercise 1.1] asks to prove that
g(x)=\frac{1}{1-x} (1)

is a monotonically increasing function over the interval 0<x<1. Furthermore it asks to show that due to the monotonicity of the peaks of the spectral estimate shown in  [1, Figure 1.2b] that they must also be present in the spectral estimate of [1, Figure 1.2a].
Solution: We notice that the first derivative of g(x) at any domain point x is positive g'(x)=\frac{1}{(1-x)^2}>0 \forall x \in  (0,1) thus g(x) is a monotonically increasing function.  For a function y=y(x) with range y\in (0,1) the extreme points (e.g.: maxima) of y(x) can be found on domain points where the first derivative is zero or at the boundaries. The function f(x)=g \circ y(x)=g(y(x)) has a derivative equal to
\frac{df}{dx}=\frac{dg}{dy}\frac{dy}{dx}
=\frac{1}{(1-y(x))^2}\frac{dy}{dx}

Thus f(x) has the same local extrema as y(x) and peaks in y(x) are shown as such in f(x) because rising  sequences in y(x) rise also in f(x) to higher values, and the first derivative of f(x) is zero at the same points where the derivative of y(x) is zero; Q.E.D..

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