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Prove that $|\liminf a_n|\geq \liminf |a_n|$ for any real sequence $\{a_n\}_{n\geq 1}$.

How can I use the results like $\liminf(-a_n)=-\limsup(a_n)$ here?

Can we apply the facts like $a_n\leq |a_n|$, $-|a_n|\leq -a_n$.

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    $\begingroup$ Apply the first, and simplify the second, using that $|a_n|=\max(a_n,-a_n)$. $\endgroup$
    – Bernard
    Oct 4 '20 at 12:05
  • $\begingroup$ Sorry Sir, I am not getting what exactly the 'first' and 'second' means. $\endgroup$ Oct 4 '20 at 12:09
  • $\begingroup$ The first & second results that you mention. $\endgroup$
    – Bernard
    Oct 4 '20 at 12:10
  • $\begingroup$ But, I am not getting the inequality yet. $\endgroup$ Oct 4 '20 at 12:15
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Choose a subsequence $a_{n_k}$ such that $\lim_k a_{n_k}= \liminf_n a_n=:L$. Then $\lim_k |a_{n_k}| = |L|$ and since the limit inferior is the smallest limit of a subsequence, we obtain $\liminf |a_n| \leq |L| = |\liminf a_n|$, proving the claim.

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