$\limsup_n a_n + \limsup_n b_n \le \limsup_n c_n + \limsup_n d_n$ if $a_n+b_n=c_n+d_n$ and $a_n$ maximum

I already posed some questions on inqualities between superior limits of real sequences, so here there is another one:

Let $(a_n)_{n\ge 1}, (b_n)_{n\ge 1}, (c_n)_{n\ge 1}$, and $(d_n)_{n\ge 1}$ be sequences of reals in $[0,1]$ for which:

(i) for all $n$, it holds $\max(a_n,b_n,c_n,d_n)=a_n$;

(ii) for all $n$, it holds $a_n+b_n=c_n+d_n$.

Question: Is it true that $$\limsup_n a_n + \limsup_n b_n \le \limsup_n c_n + \limsup_n d_n?$$

Here’s a counterexample. Let

$$a_n=\begin{cases} 1,&\text{if }n\text{ is even}\\ 1/2,&\text{if }n\text{ is odd}\;, \end{cases}$$

$$b_n=\begin{cases} 0,&\text{if }n\text{ is even}\\ 1/2,&\text{if }n\text{ is odd}\;, \end{cases}$$

and $c_n=d_n=\dfrac12$ for all $n\in\Bbb N$; clearly $a_n+b_n=1=c_n+d_n$ and $\max\{a_n,b_n,c_n,d_n\}=a_n$ for each $n\in\Bbb N$.

Then

$$\limsup_na_n+\limsup_nb_n=1+\frac12=\frac32>1=\frac12+\frac12=\limsup_nc_n+\limsup_nd_n\;.$$

• Perfect :)_____ – Paolo Leonetti Aug 8 '15 at 15:55