Cauchy sequence $x_n=\sqrt{a+x_{n-1}}$ I have to show that this sequence
$$
x_n=\sqrt{a+x_{n-1}} \hbox{ with } x_1=\sqrt{a}
$$
is a Cauchy sequence for every $a>0$. I have done the following calculations:
$$
\left| x_{n+2}-x_{n+1} \right|=\left| \sqrt{a+x_{n+1}}-x_{n+1}\right|=\left|\frac{a+x_{n+1}-x_{n+1}^2}{\sqrt{a+x_{n+1}}+x_{n+1}} \right|= \left|\frac{x_{n+1}-x_n}{\sqrt{a+x_{n+1}}+x_{n+1}}\right|
$$
I don't come up with a boundary for the denominator so that
$$
\left|\frac{x_{n+1}-x_n}{\sqrt{a+x_{n+1}}+x_{n+1}}\right|<k\cdot\left|x_{n+1}-x_n\right| \hbox{ with } k<1
$$
Any hint? I know I can use induction to easily prove that the sequence is convergent, but I'd like to prove it's Cauchy without using convergence. Thank you in advance.

Claim. The given sequence is bounded from above by $$ \frac{1 + \sqrt{4a + 1}}{2} = \alpha $$
Proof. We proceed by induction. For $ k = 1 $ we have that
$$ \sqrt{a} = \frac{2\sqrt{a}}{2} = \frac{\sqrt{4a}}{2} < \frac{1 + \sqrt{4a + 1}}{2} $$
Now, assuming that the statement is true for $ k = n $, we have
$$ x_{n+1} = \sqrt{a + x_n} < \sqrt{ \frac{1 + 2a + \sqrt{4a + 1}}{2} } $$
Observe that
$$ \left( \frac{1 + \sqrt{4a + 1}}{2} \right)^2 = \frac{4a + 2 + 2\sqrt{4a + 1}}{4} = \frac{1 + 2a + \sqrt{4a + 1}}{2} $$
which establishes the bound for $ x_{n+1} $.
Now, we prove that the sequence is monotone increasing. For this, the following inequality is crucial:
Claim. For $ 0 \leq x < \alpha $, we have that $ x < \sqrt{a+x} $.
Proof. The given inequality is true precisely when $ x^2 - x - a < 0 $. The inequality factors as $ (x - \alpha)(x - \alpha') < 0 $ where $ \alpha' < 0 $, therefore the second factor is always positive when $ x \geq 0 $, and the first factor is negative precisely when $ x < \alpha $. 
Now, since we have $ 0 \leq x_k < \alpha $ for all $ k $, it follows by this claim that
$$ x_{k+1} = \sqrt{a + x_k} > x_k $$
Since the sequence $ x_k $ is monotone increasing and bounded from above, by the monotone convergence theorem it is convergent; and thus Cauchy.
