= Solution
The <contraction mapping theorem> states: if $(X,d)$ is a nonempty complete <metric space> and $f:X\to X$ satisfies $d(fx,fy)\le qd(x,y)$ for a fixed $0\le q<1$, then it has exactly one <fixed point> and the iteration from every starting point converges to it.
To prove existence, choose $x_0$ and let $x_{n+1}=f(x_n)$. Induction gives $d(x_{n+1},x_n)\le q^nd(x_1,x_0)$. The <triangle inequality> and geometric sum then imply, for $m>n$,
$$
d(x_m,x_n)\le\frac{q^n}{1-q}d(x_1,x_0).
$$
The iterates are a <Cauchy sequence>; completeness supplies their limit $x_*$. The contraction is continuous, so $f(x_*)=\lim f(x_n)=\lim x_{n+1}=x_*$. If $y_*$ is another <fixed point>, $d(x_*,y_*)\le qd(x_*,y_*)$ forces equality of the points. The same estimate gives geometric convergence from every initial point.
For the particular map, the original PDF gives $X=[\sqrt{a/2},\infty)$, which is a closed, complete subset of the real line. For positive $x$, the arithmetic-geometric mean inequality gives $f(x)=(x+a/x)/2\ge\sqrt a$, proving $f(X)\subset X$. Furthermore
$$
f'(x)=\frac12\left(1-\frac a{x^2}\right),\qquad -\frac12\le f'(x)<\frac12\quad(x\in X).
$$
The <mean value theorem> therefore gives $|f(x)-f(y)|\le|x-y|/2$, a genuine contraction. The fixed-point equation is $x^2=a$, and only the positive root belongs to $X$, so
$$
\boxed{x_*=\sqrt a.}
$$
The printed square-root endpoint is essential: the TeX transcription loses it.
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