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?
 A: Hint:
$$x_n- x_{n+1}=x_n-\frac{1}{2}\left(x_n+\frac{a}{x_n}\right)=\frac{1}{2} \frac{x_n^2-a}{x_n} \geq 0$$ since $x_n^2 \geq a$ (?)
A: Lemma: $x_n>\sqrt{a}$ for all $n\ge 1$.
Proof: The proof is by induction. Suppose $x_n>\sqrt{a}$. Divding $a$ by both sides, you get $\frac{a}{x_n}<\sqrt{a}$. Combining these, you get $x_n>\frac a{x_n}$, so $\sqrt{x_n}>\sqrt{\frac a{x_n}}$, so 
$$
\left(\sqrt{x_n}-\sqrt{\frac a{x_n}}\right)^2>0
$$
This rearranges to 
$$
\frac12\left(x_n+\frac{a}{x_n}\right)>\sqrt{a}
$$
which is exactly $x_{n+1}>\sqrt{a}$. $\hspace{.2cm}\square$

Now we know $x_n>\sqrt{a}$ for all $n$, which as before implies $x_n>\frac{a}{x_n}$. Conclude with
$$
x_{n}=\frac12 x_n+\frac12 x_n> \frac12 x_n+\frac12\cdot \frac{a}{x_{n}}=x_{n+1} 
$$
A: I suppose $a>0$.
Now consider the function $f(x)=\frac12\left(x+\frac{a}{x}\right)$ for $x>\sqrt{a}$. Take the derivative and show that the derivative is positive then deduce that the function is an increasing one. After that start an induction. If $x_n>x_{n+1}$   then $f(x_n)>f(...$
A: Sine x1 > √a, it is a positive value, then we will have for x2:
$$x2 = \frac{1}{2}(x1 + \frac{a}{x1}) = \frac{1}{2}\frac{(x1^2+a)}{x1}$$
Now in order to compare them we have:
$$\frac{1}{2}\frac{(x1^2+a)}{x1}$$
VS
$$x1$$
Since x1 is at least √a, we substitute it in the formulas:
$$\frac{1}{2}\frac{(a^++a)}{√a^+}$$
VS
$$√a^+$$
Multiplying by 2√a^+ gives us:
$$(a^++a)$$
VS
$$2a^+$$
Obviously, the second side, x1 is bigger than x2.
For the rest, you might need to use induction approach.
A: $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
$x_{n+1}
\le x_n$.
A: Assuming $a>0$ and starting value $x_0> 0$, just apply AM-GM directly:


*

*$\frac{1}{2}\left(x +\frac{a}{x}\right) \geq \sqrt{x\cdot \frac{a}{x}}= \sqrt{a}$ with equality if and only if $x=\frac{a}{x} \Leftrightarrow x=\sqrt{a}$
We conclude for $x_0 = \sqrt{a}$ you get the constant sequence $x_n = \sqrt{a}$, otherwise for any $x_0>0, x_0 \neq \sqrt{a}$ you have $x_1 > \sqrt{a}$ and, hence $x_n > \sqrt{a}$ for $n\geq 1$.
If follows immediately 
$$\color{blue}{x_n - x_{n+1}} = x_n - \frac{1}{2}\left(x_n +\frac{a}{x_n}\right)= \frac{1}{2}\left(x_n -\frac{a}{x_n}\right) \stackrel{x_n > \sqrt{a}}{\color{blue}{>}} \frac{1}{2}\left(\sqrt{a} -\frac{a}{\sqrt{a}}\right)= \color{blue}{0}$$
