The sequences $s_{n}$ and $t_{n}$ are such that there exists an $N_{0}$ such that $s_{n}\leq t_{n}$ $\forall n>N_{0}$. I am asked to prove that $\lim \inf s_{n}\leq \lim \inf t_{n}$. Is the following correct?
Since $s_{n}\leq t_{n}$ $\forall n>N_{0}$, then $\inf\{s_{n}:n>N_{0}\}\leq \inf\{t_{n}:n>N_{0}\}$. So $\inf\{s_{n}:n>N\}\leq \inf\{t_{n}:n>N\}$ $\forall N\geq N_{0}$.
Now, since $\inf\{t_n:n>N\}$ is non-decreasing, we have two cases to consider.
First, it could be that $\inf\{t_n:n>N\}$ is unbounded, and so tends towards $+\infty$. In this case, the result follows trivially.
Second, it could be the case that $\inf\{t_{n}:n>N\}$ is bounded, and so tends towards some real number $t$.
We now show that this implies that $\inf\{s_{n}:n>N\}\leq t$, for all $N\geq N_{0}$.
Take some arbitrary $N_{1}\in\{N_{0},N_{0}+1,...\}$. We know that $\inf\{s_{n}:n>N_{1}\}\leq \inf\{t_{n}:n>N_{1}\}$. Also, since $\inf\{s_{n}:n>N\}$ is non-decreasing, $\inf\{s_{n}:n>N_{1}\}\leq \inf\{t_{n}:n>N_{1}\}\leq t$. Since $N_{1}$ was said to be any arbitrary $N\geq N_{0}$, we have that $\inf\{s_{n}:n>N\}\leq t$, for all $N\geq N_{0}$.
So then it must be that $\lim_{N\rightarrow\infty}\inf\{s_{n}: n>N\}\leq t$ [assuming the result that, if $a_{n}\leq b\in \mathbb{R}$ for all but finitely many $n$, then the limit of $a_{n}$ is also weakly less than $b$].
So $\lim_{N\rightarrow\infty}\inf\{s_{n}: n>N\}$ is also finite—call it $s$.
So, [assuming the result that, for any two sequences, if $a_{n}\leq b_{n}$ for all but finitely many $n$, $\lim a_{n}\leq \lim b_{n}$], we have that $\lim_{N \rightarrow \infty}\inf\{s_{n}:n>N\}\leq\lim_{N \rightarrow \infty}\inf\{t_{n}:n>N\}$, as desired.
Thank you.
The result that you assumed also needs the hypothesis that both sequences have a limit. Now, since both sequences $\inf \{s_n\,:\, n\geq N\}$ and $\inf \{t_n\,:\, n\geq N\}$ are increasing sequences in $N$, in particular $\inf \{t_n\,:\, n\geq N\}$ must diverge to $+\infty$, in which case the result is obvious, or converge to a real number and this limit would be also an upper bound for $\inf \{s_n\,:\, n\geq N\}$, so it also converges and the inequality follows by the result you mentioned.