complete partial order by adjoint functor theorem I cannot prove by my self the following remark

By the adjoint functor theorem for posets, having either all joins or all meets is sufficient for the other [1]

So far I got this:
Let $A$ be a poset and suppose it has all lubs, for every $X \subseteq A$ let $LX$ be the lowest upper bound. When $A$ is considered as a category, $L$ is a functor from the powerset of $A$ to $A$ (morphisms are mapped to the univeral map). Since every triangle commutes in a poset, (co)limits of (co)cones are just (lubs) glbs thus $A$ is co-complete. Also $L(\{\})$ is initial. The functor $L$ preserves colimits so it has a right adjoint $G$. Thus we have adjunction, for all $X \in \mathcal P A$, $a \in A$,
$$\frac{X \le G a}{LX \le a}$$
taking the dual of this (and using the fact the opposite in powerset is just compliment) gives
$$\frac{a^{op} \le LX^{op}}{(G a)^c \le X^c}.$$
Can I conclude from this that $A^{op}$ has colimits (lubs)? I don't know how.
Otherwise I thought of using this lemma $A$ has limits of shape $J$ iff the constant functor $\Delta : A \to [J,A]$ has a right adjoint, but I don't know how to do this argument either.
[1] http://ncatlab.org/nlab/show/complete+lattice
 A: I suppose that could work. Let me write it out cleanly. Let $A$ be a poset and let $I(A)$ be the set of all downward-closed subsets of $A$. There is then a monotone embedding $\mathop{\downarrow} (-) : A \to I(A)$ and moreover this monotone embedding preserves all meets that exist in $\mathop{\downarrow} (-)$.
Now suppose $A$ has all meets. Then the adjoint functor theorem says $\mathop{\downarrow} (-)$ has a left adjoint, say $\sup : I(A) \to A$. It is not hard to see that $I(A)$ is closed under unions (hence $I(A)$ has all joins!), and left adjoints preserve meets, so we can show that, for any subset $S \subseteq A$, 
$$\sup \bigcup_{s \in S} \mathop{\downarrow} (s) = \bigvee_{s \in S} \sup \mathop{\downarrow} (s) = \bigvee_{s \in S} s$$
hence $A$ itself has all joins.
Conversely, if $A$ has all joins, then the dual of this argument shows that $A$ has all meets.
A: In fact, one can give a much shorter argument using the adjoint functor theorem, based on your last idea (I learned of this argument in P T Johnstone's Category Theory course, so it is far from being my own idea.).
Let $A$ be a poset having all meets, i. e. limits. Then for any set $J$, the constant diagram functor $\Delta\colon A \to [J,A]$ preserves all limits, since limits in $[J,A]$ are constructed "pointwise". Thus by the adjoint functor theorem it has a left adjoint, i. e. $A$ has colimits of shape J. Therefore, $A$ has all colimits, i. e. joins.
(Note that I used index sets while talking about all colimits, not just small ones, but since we're dealing with posets, this doesn't matter.)
