Is $\liminf x_n^2\geq (\liminf x_n)^2$? If $(x_n)$ is a sequence such that $0\leq x_n\leq C$ for some $C>0$, can we conclude that
$$
\liminf x_n^2\geq  (\liminf x_n)^2 ?
$$
If so, why?
EDIT:
I think I may have found a proof. Let $a = \liminf x_n$. All we need to show is that for any $\epsilon >0$, $x_n^2>a^2-\epsilon$ eventually. To that end, let $\epsilon >0$. By the definition of $\liminf$, choose $N>0$ such that if $n\geq
 N$, then $x_n>a-\frac{\epsilon}{a+C}$. Multiplying this by $x_n\geq 0$ gives
$$
x_n^2\geq ax_n-\frac{\epsilon x_n}{a+C}>a^2-\frac{a\epsilon}{a+C}-\frac{\epsilon x_n}{a+C}\geq a^2-\epsilon
$$
as required.
 A: Let $L=\liminf x_n$, and that means there's a subsequence $x_{n_k} \to L$ as $k \to \infty$.
(i.e. $\lim_{k \to \infty} x_{n_k}=L$)
By the function $f(x)=x^2$ is a continuous, so $(\liminf_{n \to \infty} x_n)^2=L^2=(\lim_{k \to \infty} x_{n_k})^2=\lim_{k \to \infty} x_{n_k}^2= \liminf_{n \to \infty} x_n^2$.
For the last equation $\lim_{k \to \infty} x_{n_k}^2= \liminf_{n \to \infty} x_n^2$.
we can view it as $\liminf_{n \to \infty} x_n=inf(\bigcap _{N=1}^{\infty} \overline{{\{x_n|n>N\}}})$, and since $f(x)=x^2$ is continuous and order preserving, we have $(\liminf_{n \to \infty} x_n)^2=f(\liminf_{n \to \infty} x_n)=f(inf(\bigcap _{N=1}^{\infty} \overline{{\{x_n|n>N\}}}))=inf f(\bigcap _{N=1}^{\infty} \overline{{\{x_n|n>N\}}})=inf (\bigcap _{N=1}^{\infty} f(\overline{{\{x_n|n>N\}}}))=inf (\bigcap _{N=1}^{\infty} \overline{{\{f(x_n)|n>N\}}})=inf (\bigcap _{N=1}^{\infty} \overline{{\{x_n^2|n>N\}}})= \liminf_{n \to \infty} x_n^2$
A: The general result is obtained by observing that, for all $k\ge n\in \mathbb N,\   x_k\cdot y_k\ge \inf _{k\ge n}x_k\cdot \inf _{k\ge n}y_k$ 
so $\inf_{k\ge n} (x_k\cdot y_k)\ge (\inf _{k\ge n}x_k)\cdot( \inf _{k\ge n}y_k)$. 
If $x_k=y_k\ge 0$, then equality holds because the square of the infimum of any set of non-negative numbers is the infimum of the squares so
let $x'_n$ be a subsequence converging to $l=\liminf x_n$. Then,
$\liminf x^{2}_n=\lim x'^{2}_n=\lim (x'_n\cdot x'_n)=(\lim x'_n)(\lim x'_n)=l\cdot l=l^{2}=(\liminf x_n)^{2}.$
