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.

The hint.
Use for all $n\geq2$ $$s_{n+1}-\sqrt3=\frac{(s_n-\sqrt3)^2}{2s_n}>0$$ and
$$s_{n+1}-s_n=\frac{3-s_n^2}{2s_n}<0.$$
