Solution
= Solution
If $x_k\geq1$, then positivity and the <arithmetic-geometric mean inequality> give
$$
x_{k+1}=\frac12\left(x_k+\frac a{x_k}\right)
\geq\sqrt a\geq1.
$$
Thus all iterates remain in $[1,\infty)$.
Consider the strictly convex <function>
$$
f(x)=\frac{x^3}{3}-ax
$$
on $[1,\infty)$. Since $f'(x)=x^2-a$ and $f''(x)=2x$, its Newton minimization step is
$$
x-\frac{f'(x)}{f''(x)}
=\frac12\left(x+\frac ax\right).
$$
Its unique minimizer is $x^*=\sqrt a$.