Let $b_n = a_{n+1}/a_n$ for $n\ge 1$. Then $b_1=2$ and $1+b_n^3 = 2b_{n+1}$ using the given recursion. From here, we get $2(b_{n+1}-1)=b_n^3-1$, yielding \[ \frac{b_{n+1}-1}{b_n-1} = \frac{b_n^2+b_n+1}{2} \Rightarrow b_{n+1}-1 = 2^{-n}\prod_{1\le i\le n}\left(b_i^2+b_i+1\right). \]The last deduction follows from a telescoping argument. As $b_i^2+b_i+1\ge 3b_i$ by AM-GM, we get that \[ \left(\frac{2}{3}\right)^n > \frac{b_1\cdots b_n}{b_{n+1}}. \] Now, using the fact $a_k = b_{k-1}\cdots b_1$ for $k\ge 2$, we get \[ \frac{a_k^2}{a_{k+1}} = \frac{b_1\cdots b_{k-1}}{b_k},\quad\text{for}\quad k\ge 2. \]So, \[ \sum_{k=1}^N \frac{a_k^2}{a_{k+1}} = \frac12 + \sum_{2\le k\le N}\frac{b_1\cdots b_{k-1}}{b_k} = \frac12 + \sum_{1\le k\le N-1}\frac{b_1\cdots b_k}{b_{k+1}} <\frac12 + \sum_{k\ge 1}\frac{2^k}{3^k}<3, \]as desired.

Simplifying given condition implies that $$\frac{a_{n+2}}{a_{n+1}}-1=\frac{(\frac{a_2^2}{a_1^2}+\frac{a_2}{a_1}+1)...(\frac{{a_{n+1}^2}}{{a_{n}^2}}+\frac{a_{n+1}}{a_{n}}+1)}{2^n}$$ $$\text{By AM-GM } \frac{{a_{i+1}^2}}{{a_i^2}}+\frac{a_{i+1}}{a_{i}}+1 \ge \frac{3a_{i+1}}{a_i} \implies \frac{a_{n+2}}{a_{n+1}} > \frac{3^n a_{n+1}}{2^n} \implies \frac{a_{n+1}^2}{a_{n+2}} < \frac{2^n}{3^n}$$ So we need to show $$ \frac{1}{2} + \sum_{n=1}^{\infty} \frac{2^n}{3^n} < 3 \text { which is obviously true }$$

After rearranging the conditions we get: $a_{n+2}=\frac{a_{n+1}}{2}+\frac{a_{n+1}^4}{2a_n^3}$ thus the sequence is strictly increasing. Furthermore notice that: $\frac{a_n^2}{a_{n+1}}\le \bigg(\frac{4}{9}\bigg)^{n-1}$ for $n\ge 2$ Claim:$\frac{1}{n^2}\ge \bigg(\frac{4}{9}\bigg)^{n-1}$ Proof: $\Longrightarrow 9^{n-1}>8^{n-1}=4^{n-1}\cdot2^{n-1}\Longleftrightarrow 2^{n-1}\ge n^2 \implies 2^{n-1}-n^2\ge 0$ Let $f(x)=2^{x-1}-x^2 \Longrightarrow f'(x)=2^{x-1}\cdot\ln{2}-2x>2^{x-2}-2x>0$ for $x\ge6$. Thus the function is strictly increasing for $n\ge6$ $\blacksquare$ Notice that: $\sum_{i=1}^6\frac{a_i^2}{a_{i+1}}+\sum_{i=7}^n\frac{a_i^2}{a_{i+1}}\le \sum_{i=1}^6\frac{a_i^2}{a_{i+1}} +\sum_{i=7}^{\infty}\frac{1}{i^2}=\sum_{i=1}^6\frac{a_i^2}{a_{i+1}}+\frac{\pi^2}{6}-\sum_{i=1}^6\frac{1}{i^2}<\frac{7}{3}+\frac{\pi^2}{6}-1.4$ This implies that: $7+\frac{\pi^2}{2}-4.2<9\Longrightarrow \pi^2<12.4$ which is clearly true. QED And we are done!