Team Selection Test for IMO 2024

Answer: $\lambda =2^{2024}$. Let $x_i=a_i-a_{i+1}$. Since for all real numbers $$\max \{x,y\}=\frac{x+y}{2}+\left|\frac{x-y}{2}\right|$$ we have $$b_i-b_{i+1}=x_i+\frac{x_{i+1}}{2}+\frac{x_{i+2}}{2}+\left|\frac{x_{i+1}}{2}-\frac{x_{i+2}}{2}\right|(1)$$

Let $y_i=\frac{x_i}{2}+\left|\frac{x_i}{2}\right|$ and $z_i=\frac{x_i}{2}-\left|\frac{x_i}{2}\right|$. Then, $y_i=x_i$, $z_i=0$ for $x_i\ge 0$ and $y_i=0$, $z_i=x_i$ for $x_i<0$. Inserting it to (1) we get $$(b_i-b_{i+1}{)}^{2024}=(x_i+y_{i+1}+z_{i+2}{)}^{2024}\le 2^{2023}(x_i^{2024}+(y_{i+1}+z_{i+2}{)}^{2024})(2)$$ by the Power Mean Inequality. Because $y_i$ is always a non-negative real number and $z_i$ is always a non-positive real number, we have $$|y_{i+1}+z_{i+2}|\le |y_{i+1}|+|z_{i+2}|$$ which in turn implies that $$(y_{i+1}+z_{i+2}{)}^{2024}\le y_{i+1}^{2024}+z_{i+2}^{2024}$$

Since $y_i^{2024}+z_i^{2024}=x_i^{2024}$ for all indices $i$, by summing the inequalities (2) for all indices $i$ we get the desired inequality. To prove that $\lambda =2^{2024}$ is the best possible value, consider $n=2m$ and choose the sequence as $\{a_i\}=\{1,2,\dots ,n-1,n,n-1,\dots ,1\}$. Then, $\lambda$ has to satisfy the condition $$\lambda \cdot n\ge 2^{2024}(n-4)+1+1+0+0$$ for every $n$. Taking $n$ to infinity, we see that $\lambda$ should be at least $2^{2024}$.