<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] 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:
$$
b_{n+1} = b_n + \left(\frac{-1}{2}\right)^{n-1}(b_{2}-b_{1}) = b_{n-1} + \left(\left(\frac{-1}{2}\right)^{n-1}+\left(\frac{-1}{2}\right)^{n-2}\right)(b_2-b_1) = \;\cdots
$$
</s>