I am trying to find the limit of a recursively defined sequence:
$$a_{n+1}=\frac{{a_n}^2+2}{2a_n}$$ for $a_1=1$
To do this I need to first show that the sequence is monotonically decreasing so I tried to prove this using induction:
Base case: $n=1,$ $a_2 = \frac{3}{2} \ge \frac{17}{12} = a_3.$
Assume $a_{n+1} \ge a_{n+2}$ for some $n \in \mathbf{N}.$
For $n+1:$ $$a_{n+2}=\frac{{a_{n+1}}^2+2}{2a_{n+1}} \le a_{n+1}$$ $${a_{n+1}}^2+2 \le (2a_{n+1})^2 \tag{1}$$
For $a_{n+3}$ I get $$\frac{{a_{n+1}}^4+12{a_{n+1}}^2+4}{4{a_{n+1}}^3+8a_{n+1}}$$ I know I have to use $(1)$ to show $$\frac{{a_{n+1}}^2+2}{2a_{n+1}} \ge \frac{{a_{n+1}}^4+12{a_{n+1}}^2+4}{4{a_{n+1}}^3+8a_{n+1}}$$ I don't really know how to get here though.
This is an elementary solution i.e. without derivative. Rearranging the equation yields the following quadratic equation:
$$ a_{n}^{2}-2a_{n+1}a_{n}+2=0 $$
From the initial relation, it is apparent that the sequence is positive real so obviously the equation must have real solutions so the discriminant is non-negative i.e.
$$ 4a_{n+1}^{2}-8\geq 0 \phantom{x} \iff \phantom{x} a_{n+1}^{2}\geq 2 $$
From the last inequality and using the fact that $a_{n+1}\geq 0\phantom{x}\forall\phantom{x}n$:
$$ 2a_{n+1}^{2}\geq a_{n+1}^{2}+2 \phantom{x} \implies \phantom{x} a_{n+1}\geq\frac{a_{n+1}^{2}+2}{2a_{n+1}}=a_{n+2} $$
Therefore, we have shown that the sequence is decreasing starting from $a_{2}$ and it is bounded below at $\sqrt{2}$