# $\limsup_n a_n+b_n+c_n+\limsup_n b_n \le \limsup_n a_n+b_n+\limsup_n b_n+c_n$?

Let $(a_n)_{n\ge 1},(b_n)_{n\ge 1},(c_n)_{n\ge 1}$ be sequences of reals in $[0,1]$. Is it true that $$\limsup_n (a_n+b_n+c_n)+\limsup_n b_n \le \limsup_n (a_n+b_n)+\limsup_n (b_n+c_n)?$$

A counter example: For $n\in\mathbb N$ let \begin{align} a_n &= \frac{1 + (-1)^n}{2}, & b_n &= \frac{1 - (-1)^n}{2}, & c_n &= a_n. \end{align} Then, we have $$\limsup_{n \to \infty} (a_n + b_n + c_n) + \limsup_{n\to\infty} b_n = 1 + \limsup_{n \to \infty} a_n + \limsup_{n\to\infty} b_n = 1 + 1 + 1 = 3$$ and $$\limsup_{n\to\infty} (a_n + b_n) + \limsup_{n\to\infty} (b_n + c_n) = 1 + 1 = 2.$$