When does equality hold in this inequality?

Question

The following inequality can be proven as follows:

Let $n\geq3$ and $0=a_0<a_1<\dots<a_{n+1}$ such that $a_1a_2+a_2a_3+\dots+a_{n-1}a_n=a_na_{n+1}$. Show that \begin{equation*} \frac{1}{{a_3}^2-{a_0}^2}+\frac{1}{{a_4}^2-{a_1}^2}+\dots+\frac{1}{{a_{n+1}}^2-{a_{n-2}}^2}\geq\frac{1}{{a_{n-1}}^2}. \end{equation*}

Solution:

The expression on the left-hand side can be rewritten as $$ \frac{a_1^2 a_2^2}{a_1^2 a_2^2 a_3^2 - a_0^2 a_1^2 a_2^2} + \frac{a_2^2 a_3^2}{a_2^2 a_3^2 a_4^2 - a_1^2 a_2^2 a_3^2} + \cdots + \frac{a_{n-1}^2 a_n^2}{a_{n-1}^2 a_n^2 a_{n+1}^2 - a_{n-2}^2 a_{n-1}^2 a_n^2}. $$ Applying the Cauchy-Schwarz inequality then yields

$$ \begin{align*} &\frac{a_1^2 a_2^2}{a_1^2 a_2^2 a_3^2 - a_0^2 a_1^2 a_2^2} + \frac{a_2^2 a_3^2}{a_2^2 a_3^2 a_4^2 - a_1^2 a_2^2 a_3^2} + \cdots + \frac{a_{n-1}^2 a_n^2}{a_{n-1}^2 a_n^2 a_{n+1}^2 - a_{n-2}^2 a_{n-1}^2 a_n^2} \\ & \ge \frac{\left( a_1 a_2 + a_2 a_3 + \cdots + a_{n-1} a_n \right)^2}{a_1^2 a_2^2 a_3^2 - a_0^2 a_1^2 a_2^2 + a_2^2 a_3^2 a_4^2 - a_1^2 a_2^2 a_3^2 + \cdots + a_{n-1}^2 a_n^2 a_{n+1}^2 - a_{n-2}^2 a_{n-1}^2 a_n^2} \\ & = \frac{a_n^2 a_{n+1}^2}{a_{n-1}^2 a_n^2 a_{n+1}^2 - a_0^2 a_1^2 a_2^2} \ge \frac{1}{a_{n-1}^2}. \end{align*} $$

When does equality hold?

Answer 1

Deduced from the equality condition of Cauchy-Schwarz inequality, we have: $$ \forall 0 \leqslant k \leqslant n-1, \, \, a_{k+3}^2 = a_k^2 + c$$ where $c > 0$ is a constant. So the equation has 3 possiblities that depends on $n$.

For example, if $n = 3K + 1$:

$$\sum_{k=0}^{K-1} (a_0 + kc)(a_1 + kc) + (a_1 + kc)(a_2 + kc) + (a_2 + kc)(a_0 + (k+1)c) = (a_1 + Kc)(a_2 + Kc) - (a_0 + Kc)(a_1 + Kc).$$

We can solve the above equation with respect to $c$(quadratic) and see what the roots are. If one of the roots is positive then the equality holds.