<s>[INST] 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.
 [/INST] 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.
</s>