Prove the given inequality. 
Let there exists a sequence such that, $a_0=1,a_1=a$, $a_{n+1}=\left(\frac{a_n^2}{a_{n-1}^2}-1\right)a_n$.
Given $a>2$, prove the following inequality $∀ k ∈\mathbb{N}$, $\sum_{i=0}^k\frac{1}{a_i}< \frac{2+a-\sqrt{a^2-4}}{2} $.

I wrote $a_i$'s recurrence up to $7$-$8$ terms looking that it will form a general observable sequence but nothing like that, but the inequality worked well. Then I tried using Induction principles but nothing worked well.
Could anyone help me in solving the question?
 A: The sequence
$$b_n:=\frac{a_{n+1}}{a_n}$$
satisfies
$$b_0=a,\quad b_{n+1}=b_n^2-1.$$
The inequality we want is equivalent to
$$\sum_{j=0}^{k-1}\frac1{b_0b_1\dots b_j}
<\frac{a-\sqrt{a^2-4}}2.$$
It is thus sufficient to prove, by induction on $k,$ that for any sequence $(c_n)$ such that $c_0>2$ and $c_{n+1}=c_n^2-1,$
$$\sum_{j=0}^{k-1}\frac1{c_0c_1\dots c_j}
<\frac{c_0-\sqrt{c_0^2-4}}2.$$
This is true for $k=1$ since the empty sum is $0$ and $\forall c\ge2\quad c>\sqrt{c^2-4}.$
Assume now that for some $k\in\Bbb N,$ the inequality is true for a sum of $k$ terms and let us prove it for a sum of $k+1$ terms. By induction hypothesis (and since $c_0>2\Rightarrow c_1:=c_0^2-1>2$)
$$\sum_{j=0}^k\frac1{c_0c_1\dots c_j}=\frac1{c_0}\left(1+\sum_{j=1}^k\frac1{c_1\dots c_j}\right)
<\frac1{c_0}\left(1+\frac{c_1-\sqrt{c_1^2-4}}2\right),$$
so there remains to check that ($\forall c_0>2$) this RHS is $\le\frac{c_0-\sqrt{c_0^2-4}}2.$ Letting $x=c_0^2-2$ and simplifying, this is equivalent to
$$(\forall x>2)\quad\sqrt{(x+1)^2-4}>1+\sqrt{x^2-4},
$$
which is obvioulsy true.
