I'm interpreting the question as asking to prove that given an ordered field $\mathbb F$, if $\mathbb F$ has the lub property, then $\mathbb F$ has the glb property. Or equivalentely that if all non-empty bounded above subsets of $\mathbb F$ have a supremum, then all non-empty and bounded below subsets of $\mathbb F$ have an infimum.
Ordered fields are irrelevant here, this can be generalized to any poset.
Let $P$ be a poset such that each of its non-empty and bounded above subsets has a supremum.
The goal is to prove that any non-empty and bounded below subset $A$ of $P$ has an infimum.
Being a universal statement, one way to prove it is to start by taking an arbitrary non-empty and bounded below subset $A$ of $P$.
Now consider the set $\downarrow A$, where $\downarrow A:=\{p\in P\colon \forall a\in A(p\leq a)\}$, that is, consider the set the lower bounds of $A$.
The set $\downarrow A$ isn't empty because $A$ is bounded below. It is also bounded above by any element of $A$.
Hence it's possible to use the hypothesis that any non-empty and bounded above subset of $P$ has a supremum particularized to $\downarrow A$.
Let $s:=\sup\left(\downarrow A\right)$.
Claim: $s=\inf(A)$.
Proof: It suffices to prove that $s$ is a lower bound of $A$ and that $\forall p\in \downarrow A(p\leq s)$. The latter part follows immediately from the fact that $s$ is an upper bound of $\downarrow A$. The first part follows from the fact that any element of $A$ is an upperbound of $\downarrow A$ and hence greater than the supremum of $\downarrow A$ which is $s$.
Since $A$ was an arbitrary bounded below and non-empty subset of $P$, it wasproved that $P$ has the glp property.
First of all, let me note that the terms minimum, maximum, infimum and supremum do not live in a vacuum. They are properties that a certain element can have with respect to a given order. Now forget about this and let us only focus on the order $(\Bbb R; \le)$ that you seem to be interested in. Given any subset of real numbers $S \subseteq \Bbb R$, we may ask whether or not there is some $r \in \Bbb R$ such that $r$ is bigger then all the elements in $S$. If such an $r$ exists, we call it an upper bound for $S$.
For example: Let $S = (0,1)$. Every element $s \in S$ is smaller than $5$ and hence $5$ is an upper bound for $S$. However, there are smaller upper bounds. $4$ is an upper bound for $S$ as well and so is $\pi$ and $\sqrt{2}$. We may hence ask if there is a least upper bound for $S$ and in fact, we have that $1$ is an upper bound for $S$ and no $r < 1$ is an upper bound for $S$. Hence $1$ is the least upper bound for $S$, which we also call the supremum of $S$.
We designed the real numbers in such a way that once a given subset of reals $S \subseteq \Bbb R$ has any upper bound, it will always have a (unique!) least upper bound, i.e. a supremum. (A supremum is, by definition, the same as a least upper bound.) Let me stress that this is by construction of the reals, it is not true for other orders. For example, in $(\Bbb Q; \le)$, the set $\{ q \in \Bbb Q \mid q^2 < 2\}$ clearly has an upper bound, but it doesn't have a supremum.
Let us return to $(\Bbb R; \le)$. Note that $\{ x \in \Bbb R \mid 0 < x \}$ does not have an upper bound and thus does not have a supremum. Recall that any subset $S \subseteq \Bbb R$ has a supremum if and only if it has an upper bound.
So what about maxima? A maximum of a set of reals $S \subseteq \Bbb R$ is a least upper bound of $S$ that is also an element of $S$. Since suprema are unique (provided they exist), we thus have that a maxima of $S$ is also the supremum of $S$ and that $S$ has at most one maximum.
Let's have a second look at $S = (0,1)$. We already know that $S$ has $1$ as it's suprema, but $1 \not \in (0,1)$. This immediately implies that $S$ does not have a maximum. (By the argument above, if $S$ had a maximum, it would be equal to its supremum. But the supremum of $S$ is not an element of $S$ and hence not a maximum.)
I'll leave the case for infima of $S \subseteq \Bbb R$ (which are greatest lower bounds of $S$) and minima of $S$ (which are greatest lower bounds of $S$ that are also elements of $S$) to you.
Best Answer
The expression $$ \inf_{n \in \mathbf Z^+} \frac{2n^2 + 3n +1}{6n^2} $$ means to define a set $$ S = \left\{ \left. \frac{2n^2 + 3n +1}{6n^2} \right| n = 1,2,3\ldots \right\} $$ and compute $\inf S$.
It's not hard to see that $$ f(n) = \frac{2n^2 + 3n +1}{6n^2} = \frac13 + \frac{1}{2n} + \frac{1}{6n^2}, $$ so as $n$ grows, $f(n)$ slowly decreases towards $1/3$, with $$\lim_{n \to \infty} f(n) = \frac13.$$ Therefore, $\inf S = 1/3$.