" wrote: note that $ a > 2$,so there exists a positive real $ b$ such that $ a = b + \frac 1b$ hence: $ a^2 = b^2 + \frac 1{b^2} + 2\Rightarrow a^2 - 2 = b^2 + \frac 1{b^2}$ so we have: $ a_2 = a(a^2 - 2) = \left(b^2 + \frac 1{b^2}\right)\left(b + \frac 1b\right)$ $ a_3 = \left((\frac {a_2}{a_1})^2 - 2\right)a_2 = \left((b^2 + \frac 1{b^2})^2 - 2\right)a_2 = \left(b^4 + \frac 1{b^4}\right)\left(b^2 + \frac 1{b^2}\right)\left(b + \frac 1b\right)$ continuing this,we get that: $ a_n = \left(b^{2^n - 1} + \frac 1{b^{2^n - 1}}\right)\cdots \left(b^2 + \frac 1{b^2}\right)\left(b + \frac 1b\right)$ so we have: $ \sum_{i = 0}^n\frac 1{a_i} = 1 + \frac {b}{b^2 + 1} + \frac {b^3}{(b^2 + 1)(b^4 + 1)} + \ldots + \frac {b^{2^n - 1}}{(b^2 + 1)(b^4 + 1)\cdots (b^{2^n} + 1)}$ on the other hand we have: $ \frac 12\left(a + 2 - \sqrt {a^2 - 4}\right) = \frac 12\left(b + \frac 1b + 2 - (b - \frac 1b)\right) = 1 + \frac 1b$ so we have to show that: $ \frac {b^2}{1 + b^2} + \frac {b^4}{(1 + b^2)(1 + b^4)} + \ldots + \frac {b^{2^n}}{(1 + b^2)(1 + b^4)\cdots (1 + b^{2^n})} < 1$ now note that: $ \frac {b^{2^k}}{(1 + b^2)\cdots (1 + b^{2^k})} = \frac 1{(1 + b^2)\cdots (1 + b^{2^k - 1})} - \frac 1{(1 + b^2)\cdots (1 + b^{2^k})}$ now summing up this equation for $ 1\leq k\leq n$ we get that: $ \frac {b^2}{1 + b^2} + \ldots + \frac {b^{2^n}}{(1 + b^2)\cdots (1 + b^{2^n})} = 1 - \frac 1{(1 + b^2)\cdots (1 + b^{2^n})} < 1$ QED

Let $S(a)$ be the sum of the LHS of the inequaltiy when $a_1=a$ when $k$ goes to infinity. Thus, $S(a)=\frac{1}{1}+\frac{1}{a}+\frac{1}{ (a^2-2)a }+\frac{1}{((a^2-2)^2-2)(a^2-2)a}+\cdots$. We have that $a(S(a)-1)=S(a^2-2)$. Let $S(a)=f(a)+\frac{2+a-\sqrt{a^2-4}}{2}$. Subsituting and simplifying: $af(a)=f(a^2-2)$. It is clear to see that $1\le S(a)\le 2$ (since $a>2$). Thus, if $f(a)=\epsilon \neq 0$, then there exists $x$ such that $|f(x)|>2$, but that would imply that $S(a)=f(a)+\frac{2+a-\sqrt{a^2-4}}{2}$ is outside of the range of $[1,2]$, contradiction. Thus, $f(a)\equiv 0$ and $S(a)=\frac{2+a-\sqrt{a^2-4}}{2}$. Thus, all the partial sums are less than $S(a)=\frac{2+a-\sqrt{a^2-4}}{2}$, as desired.

The solution as given in Bhabak Ghalebi's post is hard to realize(probably too hard....its not an usual transormation).Are we to have no easier solutions???