limits proof by induction Let $ \{a_n\}_{n \geq 0},\{b_n\}_{n \geq 0}  $ be two sequences. The sequences are given as $ b_n:= \dfrac{a_{n-1}+b_{n-1}}2 $ $a_n = \frac{c}{b_{n}} $ for $n \geq 1$ with $b_0 = b $ and $a_0 = \dfrac{c}{b}$.
I know that: $$ \frac{a_{0}+b_{0}}{2} > \sqrt{a_{0}b_{0}} $$
Now I want to prove that for all integers n: $$ a_n < \sqrt{c} < b_n  $$
I wanna prove this by induction over $n$.
For the base case: $n=1$
$$ a_1 = \frac{c}{b_{1}} = \frac{c}{0.5(a_{0}+b_{0})} < \frac{c}{\sqrt{a_{0}b_{0}}} = \frac{c}{\sqrt{c}} = \sqrt{c} = \sqrt{a_{1}b_{1}} < {0.5(a_{1}+b_{1})} = {0.5(\frac{c}{b_1}+b_{1})} = {0.5(\sqrt{c}+b_{1})}  $$
But for me it seems to go nowhere :/
Can somebody help me? :)
 A: There are several issues with your question. First, you should require $b \gt 0$ and $c \ge 0$ for it to work properly (e.g., if $c \lt 0$, then $\sqrt{c}$ is not even a real number). Second, your $\lt$ should be $\le$, e.g., if $b = c = 1$. Third, what you want to prove for "all integers $n$" should be for positive integers $n$ since your relation doesn't necessarily hold for $n = 0$ (e.g., $b = 1$ and $c = 2$).
This means you want to prove that for all integers $n \ge 1$ that
$$a_n \le \sqrt{c} \le b_n \tag{1}\label{eq1A}$$
Note you can prove this directly. However, to do it using induction, start with your base case of $n = 1$. As you stated, using the inequality of arithmetic and geometric means gives
$$b_1 = \frac{a_{0} + b_{0}}{2} \ge \sqrt{a_{0}b_{0}} = \sqrt{c} \tag{2}\label{eq2A}$$
Using $a_1 = \frac{c}{b_1} \implies a_1 b_1 = c$ and multiplying both sides above by $a_1$ (note $a_1 \ge 0$ so the inequality doesn't change) gives, for $c \gt 0$, that
$$\begin{equation}\begin{aligned}
a_1 b_1 & \ge a_1\sqrt{c} \\
c & \ge a_1\sqrt{c} \\
\sqrt{c} & \ge a_1
\end{aligned}\end{equation}\tag{3}\label{eq3A}$$
Note if $c = 0$, then $a_n = 0$ for all $n \ge 0$ so \eqref{eq3A} (and \eqref{eq5A} below) still hold. This confirms the base case.
For the inductive step, assume \eqref{eq1A} is true for $n = k$ for some integer $k \ge 1$. Since $a_k b_k = c$, using the AM-GM inequality again gives, similar to \eqref{eq2A},
$$b_{k+1} = \frac{a_{k} + b_{k}}{2} \ge \sqrt{a_{k}b_{k}} = \sqrt{c} \tag{4}\label{eq4A}$$
Basically repeating the procedure used with \eqref{eq3A} gives
$$\begin{equation}\begin{aligned}
a_{k+1}b_{k+1} & \ge a_{k+1}\sqrt{c} \\
c & \ge a_{k+1}\sqrt{c} \\
\sqrt{c} & \ge a_{k+1}
\end{aligned}\end{equation}\tag{5}\label{eq5A}$$
This shows \eqref{eq1A} is also true for $n = k + 1$ so, by induction, it's true for all $n \ge 1$.
