We are given a sequence
$a_1=1,\ a_{n+1}=\frac{a_n^2+c}{2a_n}$ for all $n\geq 1\ (c>0)$
and we need to prove it's defined for all $n$ and converges (and to what value, but that's easy).
My attempt: I want to show that $a_n$ is bounded and thus is defined. I showed that for all $n\geq 2,\ a_n\geq \sqrt c$ (the limit of the sequence) so $a_n\geq \text{min}(1,\sqrt c)$, and I'm pretty sure $a_2=\frac{1+c}{2}$ is an upper bound, but I cannot show it really is/find another easier to prove upper bound.
I feel like I'm the the wrong direction, any help is welcomed!
Answer
Let $f(x)=\frac{1}{2}\left(x+\frac{c}{x}\right)$ (I guess that $c>0$).
i) Note that if $x>0$ the $f(x)>0$, which means that $a_n>0$ (the sequence is well-defined).
ii) For $x>0$, $f(x)\leq x$ iff $x\geq \sqrt{c}$ (why),
which means that if $a_n\geq \sqrt{c}$ then $a_{n+1}=f(a_n)\leq a_n$.
iii) If $x\geq \sqrt{c}$ then $f(x)\geq \sqrt{c}$ (why?) which means that
if $a_n\geq \sqrt{c}$ then $a_{n+1}=f(a_n)\geq \sqrt{c}$.
Now $a_2=\frac{1+c}{2}\geq \sqrt{c}$, therefore, by ii) and iii), for $n\geq 2$, $a_n\in [\sqrt{c},a_2]$ and $a_n$ is decreasing.
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