${a_n} \ge {b_n}\forall n \in $
Prove: $\liminf({a_n}) \ge \liminf({b_n})$
I proved it by contradiction. Let's assume $\liminf({a_n}) < \liminf({b_n})$.
$a := \liminf({a_n})$
$b := \liminf({b_n})$
So, by the definition of partial limit:
$\eqalign{
& \forall \varepsilon > 0\forall {N_0} \in \exists n > {N_0}st|{a_n} - a| < \varepsilon \cr
& \forall \varepsilon > 0\forall {N_0} \in \exists n > {N_0}st|{b_n} - b| < \varepsilon \cr} $
Then,
$\eqalign{
& a - \varepsilon < {a_n} < a + \varepsilon \cr
& b - \varepsilon < {b_n} < b + \varepsilon \cr} $
Lets substract the two inequalities:
$a_n-b_n < a-b < 0$
$a_n-b_n$ must be positive by definition, and so the last line is a contradiction.
Therefore, liminf of $a_n$ must be equal/greater than liminf of $b_n$.
Does the logical statements are good enough?
Do you think it can be polished?
Since $\liminf a_{n} = a$ it follows from definition of "$\liminf$" that for any $\epsilon > 0$ we have $a_{n} > a - \epsilon$ for all sufficiently large $n$ and $a_{n} < a + \epsilon$ for an infinity of values of $n$. Similarly we have a statement for $b_{n}$ and $b$. Assume on the contrary that $a < b$ and we set $2\epsilon = b - a$ so that $b - \epsilon = a + \epsilon$. Now we can see that $b_{n} > b - \epsilon = a + \epsilon$ for all sufficiently large values of $n$. This means that $a_{n} \geq b_{n} > a + \epsilon$ for all sufficiently large values of $n$ and this contradicts the statement "$a_{n} < a + \epsilon$ for an infinity of values of $n$". This contradiction shows that $a \geq b$.
Note that even if $a_{n} > b_{n}$ for all $n$ we can only infer $a \geq b$ and not $a > b$. In fact we can have $a = b$ in this case too.