Will 28points be enough for gold?

Acorn-SJ wrote: $$\frac{a_{n}}{b_{n}}+\frac{b_{n}}{a_{n}} \le M + \frac{1}{M}=\frac{M^{2}+1}{M}$$ Could you explain this better?

Let ${P_{n}}={a_{n}b_{n}}$ for all positive integers $n$. Then $$a_{n+1}b_{n+1}= a_n^2 + b_n^2=P_{n+1}$$$$a_nb_n=a_{n+1}+b_{n+1}=P_n$$Let $\frac{P_{n+1}}{P_{n}}=k_{n}$ for all positive integers $n$. By AM-GM, $$P_{n+1}=a_n^2 + b_n^2\ge 2a_nb_n=2P_{n}$$Hence $$k_n\ge 2$$However, by using the two given equations, $$P_{n+1}=P_{n-1}^2-2P_n$$Dividing $P_n$ of the both sides of the above eqation, $$\frac{P_{n+1}}{P_n}=\frac{P_{n-1}^2}{P_n}-2$$$$\implies k_n=\frac{P_{n-1}^2}{P_{n-2}^2-2P_{n-1}}-2>\frac{P_{n-1}^2}{P_{n-2}^2}-2=k_{n-2}^2-2$$So we got $$k_n>k_{n-2}^2-2$$By subtracting $k_{n-2}$ from the above inequallity, $$k_n-k_{n-2}>k_{n-2}^2-k_{n-2}-2$$Let $k_n=2+\epsilon_n$ for all positive integers $n$. For all integers $n\ge3$, $$k_n>k_{n-2}^2-2\ge2^2-2=2$$Hence $\epsilon_n>0$ for all integers $n\ge3$. However, for all integers $m\ge2$, $$2+\epsilon_{2m+1}=k_{2m+1}>k_{2m-1}^2-2=(2+\epsilon_{2m-1})^2-2=\epsilon_{2m-1}^2+4\epsilon_{2m-1}+2>2+4\epsilon_{2m-1}$$Hence $$\epsilon_{2m+1}>4\epsilon_{2m-1}$$Therefore, $$\lim_{m\rightarrow\infty}\epsilon_{2m-1}=\infty$$For all integers $m\ge2$, $$k_{2m+1}-k_{2m-1}>k_{2m-1}^2-k_{2m-1}-2=(2+\epsilon_{2m-1})^2-(2+\epsilon_{2m-1})-2=\epsilon_{2m-1}^2+3\epsilon_{2m-1}>\epsilon_{2m-1}$$Therefore, $$\lim_{m\rightarrow\infty}(k_{2m+1}-k_{2m-1})=\infty$$So there exists an odd integer $i$ such that $i>1$, $k_{i}>2023^{2023}+1$. Hence $$\frac{a_i}{b_i}+\frac{b_i}{a_i}=\frac{a_i^2+b_i^2}{a_ib_i}=\frac{P_{i+1}}{P_i}=k_i>2023^{2023}+1$$$$\implies \frac{a_i}{b_i}>2023^{2023}+1-\frac{b_i}{a_i}\ge2023^{2023}+1-1=2023^{2023}$$Therefore, $$\frac{a_i}{b_i}>2023^{2023} \blacksquare$$