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**ThePerfectHacker** We will assume $\displaystyle n\geq 2$ and so $\displaystyle s_n = \tfrac{1}{2}(s_{n-1} + xs_{n-1}^{-1})$. Therefore, $\displaystyle s_n^2 = \tfrac{1}{4}(s_{n-1}^2 + 2x + x^2s_{n-1}^{-2})$. We get from here $\displaystyle s_n^2 = x + (s_{n-1} - xs_{n-1})^2 \geq x \implies s_n \geq \sqrt{x}$. This was your first half of your problem.

Now, $\displaystyle s_{n+1} - s_n = \tfrac{1}{2} (s_n + xs_{n}^{-1}) - s_n = \tfrac{1}{2}(xs_n^{-1} - s_n) < \tfrac{1}{2}(s_n - s_n) = 0$.

(The last inequality follows because $\displaystyle s_n^2 > x \implies s_n > xs_n^{-1}$)