Monotonic recursive sequence with recursive term in denominator: $s_{n+1} = \frac{1}{2} \left(s_n + \frac{3}{s_n}\right)$ I am trying to show that $s_{n+1} = \frac{1}{2} \left(s_n + \frac{3}{s_n}\right)$ is a monotonic decreasing sequence for $n \ge 2$. Currently, my approach using induction is stuck because of the $\mathbf{s_n}$ term appearing in the denominator. 
What I have so far:
Calculating a few terms: $s_1 = 1$, $s_2 = 2$, $s_3 = \frac{7}{4} < s_2$.
For this to be monotonic decreasing, we have to show $s_{k+2} < s_{k+1}$. 
For the base case, we assume for $k$ we have $s_{k+1} < s_k.$
The inductive step: to show $s_{k+2} < s_{k+1}$, i.e. to show $$\color{grey}{s_{k+2} = }\frac{1}{2} \left(s_{k+1} + \frac{3}{s_{k+1}}\right) < \frac{1}{2} \left(s_{k} + \frac{3}{s_{k}}\right) \color{grey}{= {s_{k+1}}_.}$$
In order to show $\dfrac{1}{2} \left(s_{k+1} + \frac{3}{s_{k+1}}\right) < \dfrac{1}{2} \left(s_{k} + \frac{3}{s_{k}}\right)$, I am stuck:
Given, $s_{k+1} < s_k$, I cannot then say $s_{k+1} + \mathbf{\frac{3}{s_{k+1}}} < s_k + \mathbf{\frac{3}{s_{k}}}$, because the $s_k, s_{k+1}$ terms which appears in the denominator may reverse the inequality. Any pointers on how to proceed further?
Disclaimer: I am revising real analysis on my own from the Kenneth Ross book, this is not strictly homework.

Assume $s>0$ and notice that for $s\ne\sqrt3$
$$\frac12\left(s+\frac3s\right)>\sqrt3$$ as can be established by computing the minium.
So for $n>1$, we have
$$s_n>\sqrt3\implies s_n^2>3\implies s_n>\frac3{s_n}\implies s_n>\frac12\left(s_n+\frac3{s_n}\right)=s_{n+1}.$$

At the same time this establishes that the sequence is decreasing and bounded below by $\sqrt3$, thus is convergent.
