Equivalent to $\frac{1}{a_{n+1}a_{n+2}}\ge \frac{1}{2023}+\frac{1}{a_{n}a_{n+1}}$, rest is easy.

it is equivalent $a_0>a_2>.....$ and $a_1>a_3>....$ 4 cases are remain and reader can do it him/her self. 1.$a_0 <1$ $a_1<1$ 2.$a_0 <1 $ $a_1 \geq 1$ 3.$a_0 \geq 1$ $a_1 \geq 1$ 4.$a_0 \geq 1$ $a_1 <1$