# How to prove these inequalities: $\liminf(a_n + b_n) \leq \liminf(a_n) + \limsup(b_n) \leq \limsup(a_n + b_n)$ [duplicate]

The inequalities are:

$$\liminf(a_n + b_n) \leq \liminf(a_n) + \limsup(b_n) \leq \limsup(a_n + b_n)$$

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## marked as duplicate by Martin Sleziak, amWhy, Mice Elf, voldemort, drhabAug 29 '14 at 17:42

Proofs like this are usually easiest done indirectly: Assume "$>$" holds instead of "$\le$". Then the difference is some $\epsilon>0$, ... –  Hagen von Eitzen Oct 1 '12 at 9:47
Hint: First show that $\limsup$ is subadditive, that is, $\limsup(a_n + b_n) \le \limsup a_n + \limsup b_n$ for real sequences $(a_n)$, $(b_n)$. From this conclude using $-\limsup(-a_n) = \liminf a_n$ that $\liminf$ is superadditive (inequality $\ge$ in the above). Then you can use all this to prove $\liminf (a_n + b_n) - \limsup b_n = \liminf (a_n + b_n) + \liminf(-b_n) \le \liminf a_n$ and the other inequality you need.