Let $a_{1}>0,a_{2}>0$ and $a_{n}=\frac{2a_{n-1}a_{n-2}}{a_{n-1}+a_{n-2}}, n>2$, then $\{ a_{n}\}$ converges to $\frac{3a_{1}a_{2}}{a_{1}+a_{2}}$. Let $a_{1}>0,a_{2}>0$ and $a_{n}=\dfrac{2a_{n-1}a_{n-2}}{a_{n-1}+a_{n-2}}, n>2$, then $\{ a_{n}\}$ converges to $\dfrac{3a_{1}a_{2}}{a_{1}+a_{2}}$.

My attempt:
  \begin{align}
  a_{n} &= \frac{2a_{n-1}a_{n-2}}{a_{n-1}+a_{n-2}} \\
  &= \frac{2}{\dfrac{1}{a_{n-1}}+\dfrac{1}{a_{n-2}}} \\
 & \le \frac{1}{\sqrt{a_{n-1}a_{n-2}}}
\end{align}

I used AM- GM inequality here. I am not able to proceed further.  How to solve the question? Please help me.
 A: You are close.
From
$a_{n}
=\dfrac{2a_{n-1}a_{n-2}}{a_{n-1}+a_{n-2}}
$,
$\dfrac1{a_{n}}
=\dfrac{a_{n-1}+a_{n-2}}{2a_{n-1}a_{n-2}}
=\dfrac12(\frac1{a_{n-2}}+\frac1{a_{n-1}})
$.
Letting
$\dfrac1{a_n}
=b_n
$,
this becomes
$b_n
=\frac12(b_{n-1}+b_{n-2})
$.
You should be able to solve this.
(I think I did in
one of my answers.)
A: Hint:  let $b_n=1/a_n$ then $b_{n+1}=(b_n+b_{n-1})/2 \iff (b_{n+1}-b_n)=-(b_n-b_{n-1})/2$. Then:
$$
b_{n+1}-b_n=\frac{-1}{2}(b_n-b_{n-1})=\left(\frac{-1}{2}\right)^2(b_{n-1}-b_{n-2}) = \cdots = \left(\frac{-1}{2}\right)^{n-1}(b_{2}-b_{1})
$$
Next, telescope:
$$
b_{n+1} = b_n + \left(\frac{-1}{2}\right)^{n-1}(b_{2}-b_{1}) = b_{n-1} + \left(\left(\frac{-1}{2}\right)^{n-1}+\left(\frac{-1}{2}\right)^{n-2}\right)(b_2-b_1) = \;\cdots
$$
A: Following in the footsteps of @martycohen elsewhere on this page, a generalized version of the sequence in the OP can be constructed. Thus, consider
$$g_n=\frac{ag_{n-1}g_{n-2}}{bg_{n-1}+cg_{n-2}}=\frac{a}{\frac{b}{g_{n-2}}+\frac{c}{g_{n-1}}}$$
The let $f_n=1/g_n$ to find that
$$f_n=\left(\frac{c}{a} \right)f_{n-1}+\left(\frac{b}{a} \right)f_{n-2}$$
which I call a generalized Fibonacci sequence and have demonstrated solutions for arbitrary constants and initial conditions in various pages in MSE. See here, for example.
