Let $x_{n+1} = \frac{1}{2}(x_n + \frac{a}{x_n})$. Prove that $x_{n+1} < x_{n}$ Let $$x_{n+1} = \frac{1}{2}(x_{n} + \frac{a}{x_{n}})$$ 
Prove that $x_{n+1} < x_{n}$ for $a \geq 0$. 
Hint: Let the initial guess satisfy $x_{1} > \sqrt{a}$
I am stuck at how to begin this. I would like to use an induction proof, but there is no simple way for me to relate the base case and begin. That is I can't even establish:  $$x_{2} < x_{1}$$
How would I do this since there is no given initial term? Or is that a mistake and an initial term should be present?

$x_{n+1} 
= \frac{1}{2}(x_{n} + \frac{a}{x_{n}})
$
so
$x_{n+1}^2 
= \frac14(x_{n}^2+2a + \frac{a^2}{x_{n}^2})
$
so
$x_{n+1}^2-a 
= \frac14(x_{n}^2+2a + \frac{a^2}{x_{n}^2})-a
= \frac14(x_{n}^2-2a + \frac{a^2}{x_{n}^2})
= \frac14(x_{n}-\frac{a}{x_{n}})^2
$.
Therefore
$x_{n+1}^2 \ge a 
$.
Now we can compare
$x_{n+1}$ and
$x_m$.
$x_{n+1}^2-x_n^2
= \frac14(-3x_{n}^2+2a + \frac{a^2}{x_{n}^2})
= -\frac14(3x_{n}^2-2a - \frac{a^2}{x_{n}^2})
$.
Since,
for $n \ge 2$,
$x_n^2 \ge a$,
$\frac{a^2}{x_{n}^2}
\le a$
so
$2a + \frac{a^2}{x_{n}^2}
\le 3a
\le 3x_n^2
$
so
$3x_{n}^2-2a - \frac{a^2}{x_{n}^2}
\ge 0
$
so
$x_{n+1}^2-x_n^2
\le 0
$
so
$x_{n+1}
\le x_n$.
