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

Question

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 $a_n\leq |a_n|$, $-|a_n|\leq -a_n$.

Answer 1

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.