It's not hard to show that $a_n=\frac{1}{2^n-1}$ for every $n\in \mathbb{Z}^+$. So it's enough to show that $\sum_{i=1}^{\infty}{\frac{1}{2^i-1}} \leq \frac{34}{21}$, which, as far as I know, is true by Erdös-Borwein constant. I'm still looking for elementary bound of that inequality.

ManuelKahayon wrote: A sequence \(\{a_n\}\) is defined as follows: \(a_1 = 1\), \(a_2 = \frac{1}{3}\), and for all \(n \geq 1,\) \(\frac{(1+a_n)(1+a_{n+2})}{(1+a_n+1)^2} = \frac{a_na_{n+2}}{a_{n+1}^2}\). Prove that, for all \(n \geq 1\), \(a_1 + a_2 + ... + a_n < \frac{34}{21}\). From the initial relationship, we have: \[\frac{(a_{n+2}+1)a_{n+1}}{(a_{n+1}+1)a_{n+2}}=\frac{(a_{n+1}+1)a_{n}}{(a_{n}+1)a_{n+1}}=\frac{(a_{2}+1)a_{1}}{(a_{1}+1)a_{2}}=2\]Then, we have: \[\frac{a_{n+1}+1}{a_{n+1}}=\frac{2(a_{n}+1)}{a_{n}}\]meaning: \[\frac{a_{n}+1}{a_{n}}=2^n\]we get: \[a_{n}=\frac{1}{2^n-1}\]Finally: \[\sum_{i=1}^{n} a_{i}=\sum_{i=1}^{n} \frac{1}{2^n-1}=1+\frac{1}{3}+\frac{1}{7}+\sum_{k=4}^{n} \frac{1}{2^k-1} \leq \frac{31}{21}+\sum_{k=4}^{n}(\frac{1}{2^{k-1}-1}-\frac{1}{2^{k}-1})=\frac{31}{21}+\frac{1}{7}-\frac{1}{2^n-1}=\frac{34}{21}-\frac{1}{2^n-1} < \frac{34}{21}\]