In fact, it is true if we replace the 2 with any number greater than $2\cos\left(\frac{\pi}{8}\right)\approx1.84775907$. We prove here that $a_{i-1}+a_{i+1}<a\cdot a_i$ where $a>2\cos\left(\frac{\pi}{8}\right)$ for some $i$. Calculate $b_k=\frac{a_k}{\sin\left(\frac{\pi}{8}(k-1)\right)}$ for $k=2, 3, 4\ldots 8$, and let $b_1=b_2$ and $b_9=b_8$. Choose $\max(b_2, b_3, b_4\ldots b_8)$, say $\max(b_2, b_3\ldots b_8)=b_m$. So we have $a_m\cdot a=a\sin\left(\frac{\pi}{8}(m-1)\right)b_m$ $>2\cos\left(\frac{\pi}{8}\right)\sin\left(\frac{\pi}{8}(m-1)\right)b_m$ $=\left(\sin\left(\frac{\pi}{8}m\right)+\sin\left(\frac{\pi}{8}(m-2)\right)\right)b_m$ $\geq\sin\left(\frac{\pi}{8}m\right)b_{m+1}+\sin\left(\frac{\pi}{8}(m-2)\right)b_{m-1}$ $=a_{m+1}+a_{m-1}$. In conclusion we have $a_m\cdot a>a_{m+1}+a_{m-1}$. If we let $a=2\cos\left(\frac{\pi}{8}\right)=\sqrt{2+\sqrt{2}}$, then we have equality with the sequence $0, \sqrt{2-\sqrt{2}}, \sqrt{2}, \sqrt{2+\sqrt{2}}, 2, \sqrt{2+\sqrt{2}}, \sqrt{2}, \sqrt{2-\sqrt{2}}, 0$. In this case we have $a_{i-1}+a_{i+1}=a\cdot a_i$ for all $i=2, 3, 4\ldots 8$.

Proving it for ($a=$) $2$ is trivial; the above solution, finding the best constant, might have something to do with the Chebyshev's polynomials ...

I'm not entirely sure how to make that connection; the main result here is that $\sin a+\sin b=2\cos\left(\frac{a-b}{2}\right)\sin\left(\frac{a+b}{2}\right)$, involving both sine and cosine. It's true that the numbers in the sequence are in fact multiples of $\sin\left(\frac{\pi}{k}\right)$, but the required ordering makes it unlikely that Chebyshev's polynomials come into play here. I might be wrong; if you've worked on this idea for a bit and have made progress, you should post it.