Proving convergence and finding limit of two sequences defined indutively 
Consider the sequences $(a_n),(b_n)$ defined indutively as: $a_1>0, b_1>a_1, a_{n+1} = \dfrac{2a_nb_n}{a_n+b_n}, b_{n+1} = \dfrac{a_n +
 b_n}{2}, n \geq 1$.
Show that both sequences converge and have the same limit.

Hint: Show that $a_n < a_{n+1} < b_{n+1} < b_{n}$
I tried to prove the hint by induction:
If, by contradiction, $a_2 \leq a_1$, then:
$\dfrac{2a_1b_1}{a_1+b_1} \leq a_1 \implies 2a_1b_1 \leq a_1^2 + a_1b_1  \implies a_1b_1 \leq a_1^2 \implies b_1 \leq a_1 \perp$ 
So $a_2 > a_1$.
If, by contradiction, $b_2 \leq a_2$, then:
$\dfrac{a_1+b_1}{2} \leq \dfrac{2a_1b_1}{a_1+b_1} \implies (a_1+b_1)^2 \leq 4a_1b_1 \implies a_1^2 + 2a_1b_1 + b_1^2 \leq 4a_ab_1 \implies a_1^2 - 2a_1b_1 + b_1^2 \leq 0 \implies (a_1-b_1)^2 \leq 0 \implies a_1 = b_1 \perp$.
So $b_2 > a_2$.
If, by contradiction, $b_1 \leq b_2$, then:
$b_1 \leq \dfrac{a_1+b_1}{2} \implies 2b_1 \leq a_1+b_1 < 2b_1 \perp$.
So $b_1>b_2$, then $a_1<a_2<b_2<b_1$ and the basis of induction holds.
However I am failing at the inductive step. Also, I would like clarification on why this hint is useful. I think it leads to squeeze theorem, but I can't properly see why.
Thanks.
 A: I wouldn't use induction. Note that
$$a_{n + 1} = \frac{2a_n b_n}{a_n + b_n} = \frac{a_n b_n}{b_{n + 1}} \implies a_{n+1}b_{n+1} = a_n b_n.$$
Note also that
\begin{align*}
a_{n+1} \le b_{n+1} &\iff \frac{2a_n b_n}{a_n + b_n} \le \frac{a_n + b_n}{2} \\
&\iff 4a_n b_n \le (a_n + b_n)^2 = a_n^2 + 2a_n b_n + b_n^2 \\
&\iff (a_n - b_n)^2 \ge 0,
\end{align*}
which is true. Since this is the case, we see that
$$b_{n + 1} = \frac{a_n + b_n}{2} \le \frac{b_n + b_n}{2} = b_n.$$
Finally, we have
$$a_n b_n = a_{n+1} b_{n+1} \le a_{n+1} b_n \implies a_n \le a_{n+1},$$
giving us
$$a_n \le a_{n+1} \le b_{n+1} \le b_n.$$
This means $(a_n)$ and $(b_n)$ are respectively monotone increasing/decreasing and bounded above/below. Therefore, they must converge to respective limits $L_a$ and $L_b$. But, using the algebra of limits,
$$L_b = \lim_{n \to \infty} b_{n+1} = \lim_{n \to \infty} \frac{b_n + a_n}{2} = \frac{L_b + L_a}{2} \implies L_b = L_a.$$
A: Note that $a_n, b_n> 0$ for all $n$ and
$$
a_{n+1} = \frac {2a_n b_n} {a_n + b_n} \leqslant \frac {a_n + b_n} 2 = b_{n+1},
$$
since 
$$
(a_n + b_n)^2 = (a_n - b_n)^2 + 4a_n b_n \geqslant 4a_n b_n,
$$
where the $=$ holds iff $a_n = b_n$. Thus for inductive step, since $a_n < b_n$, $a_{n+1} < b_{n+1}$. Also, 
$$
a_{n+1} = \frac 2 {\dfrac 1{a_n} + \dfrac 1{b_n}} > \frac 2 {\dfrac 2{a_n}} = a_n
$$
and
$$
b_{n+1} = \frac {a_n +b_n}2 < \frac {b_n + b_n}2 = b_n, 
$$
hence complete the inductive step. 
Now let the limits be $a,b$ respectively, then $b = (a+b)/2$ i.e. $a=b$. This complete the proof. 
