The answer is $\lambda = 2^{2024}$. Let $\max \{x,y\} = \frac{x+y}{2} + | \frac{x-y}{2} |$ for every real $x,y$. If we denote $a_i - a_{i+1} = x_i$ , then we have $$b_i - b_{i+1} = x_i + \frac{x_{i+1}}{2} + \frac{x_{i+2}}{2} + \left| \frac{x_{i+1}}{2} \right| - \left| \frac{x_{i+2}}{2} \right|.$$Let $y_i = \frac{x_i}{2} + | \frac{x_i}{2} |$ ve $z_i = \frac{x_i}{2} - | \frac{x_i}{2} |$ for every $i$. Then we have, $y_i = x_i$, $z_i=0$; when $x_i \geq 0$ and $y_i = 0$, $z_i = x_i$ when $x_i < 0$. If we write that to the given inequality, we get, $$(b_i - b_{i+1})^{2024} = (x_i + y_{i+1} + z_{i+2})^{2024} \leq 2^{2023} \left( x_i^{2024} + (y_{i+1} + z_{i+2})^{2024} \right). $$Because of $y_i$ is always a non-negative real number and $z_j$ is always an non-positive real number, we have $$|y_{i+1} + z_{i+2}| \leq |y_{i+1}| + |z_{i+2}| \Rightarrow (y_{i+1} + z_{i+2})^{2024} \leq y_{i+1}^{2024} + z_{i+2}^{2024}.$$Since $y_i^{2024} + z_i^{2024} = x_i^{2024}$ is true for all $i$, if we sum op all these inequalities for all $i$'s we get the required inequality. For the equality case, take firs $m$ $x_i$'s as 1, a few $0$ in the middle, and then last $m$ to be $-1$, and take both $m$ and $n$ to the infinity, we can see that any number smaller than $2^{2024}$ will not satisfy the equality.