A problem on Cauchy sequences Let $\langle x_n\rangle $ be a sequence defined recursively by $ 0<a \le x_1 \le x_2\le b $  and $ x_{n+2} =\sqrt{x_nx_{n+1}} $ for each n $ \in \Bbb N $ show that $|x_{n+2} -x_{n+1}| \le \frac {b}{b+a} |x_n -x_{n+1}|$ Deduce that $\langle x_n \rangle$ is  Cauchy and find its limit.
Can some one help me on this especially in deducing its Cauchy! I think the 1st part can be proven by induction though I had some problem there too.

Well, this is not an answer, just an alternative approach.
Let $\theta_n = \ln x_n$. Then we have $\theta_{n+2} = \frac{1}{2} ( \theta_{n+1} + \theta_n)$. It should be clear from this formula that $\theta_n \in [\theta_0, \theta_1]$ for all $n$ (with appropriate adjustments for the relative order of $\theta_0, \theta_1$), so the sequence is well defined.
Then it is easy to compute that $\theta_{n+2}-\theta_{n+1} = \frac{1}{2}(\theta_n - \theta_{n+1})$, which gives $|\theta_{n+2}-\theta_{n+1}| = \frac{1}{2}|\theta_{n+1} - \theta_n|$, and so $|\theta_{n+1} - \theta_n| = \frac{1}{2^n} |\theta_2-\theta_1|$. Then summing gives (assuming $n\ge m$ for simplicity)
$$
|\theta_{n}-\theta_{m}| =|\theta_{n}-\theta_{n-1}| + ... + |\theta_{m+1}-\theta_{m}| = \frac{2}{2^m} ( 1-\frac{1}{2^{n-m}} )|\theta_2-\theta_1| \le \frac{2}{2^m} |\theta_2-\theta_1|
$$
It follows that $\theta_k$ is Cauchy, and since $x \mapsto e^x$ is uniformly continuous on $[\theta_0, \theta_1]$, the sequence $x_k$ is also Cauchy.
