Does ZFC + the Axiom of Constructibility imply the nonexistence of inaccessible cardinals? In ZFC set theory, does adding Constructibility imply the nonexistence of inaccessible cardinals like it does the nonexistence of other types of higher cardinals? 
 A: Not unless they are inconsistent altogether.
Inaccessible cardinals, Mahlo cardinals, weakly compact cardinals, and in fact much more, are all downwards absolute to $L$, in the sense that if $\kappa$ is inaccessible, then it is inaccessible in $L$, and same for Mahlo or weakly compact and so on.
In fact, many equiconsistency results utilize this, for example: if $\omega_2$ has the tree property, that implies that $(\omega_2)^V$ is weakly compact in $L$.
A: This should really be a comment, but it's too long.
As Asaf said, any inaccessible cardinal is also inaccessible in $L$, hence the existence of an inaccessible cardinal is consistent with ZFC+V=L iff it is consistent with ZFC. The idea "V=L disproves large cardinals" (which you may have heard) is true in a sense, but without context it's highly misleading.
To try to clear things up, let me briefly explain why by contrast the existence of a measurable cardinal does contradict V=L:
First of all, note that being a measurable cardinal is a more complicated property than being an inaccessible cardinal. $\kappa$ is measurable if there exists some $\mathcal{U}$ such that [stuff]. Just because such a $\mathcal{U}$ exists doesn't at first glance imply that there is such a $\mathcal{U}$ in the specific substructure $L$. By contrast,  inaccessibility has no such "witness requirement." Precisey: inaccessibility, Mahlo-ness, etc. are $\Pi_1$ properties, and all such properties are absolute from $V$ to $L$. Measurability on the other hand is $\Sigma_2$.
Now this doesn't show that measurables are inconsistent with V=L, just that the argument for consistency that we used for inaccessibles doesn't apply here. To show that measurables are in fact inconsistent with V=L, we need a more complicated argument. Roughly, we say:


*

*Let $\kappa$ be the least measurable cardinal. (Exercise: a countably closed ultrafilter on the least measurable $\kappa$ is in fact $\kappa$-closed. This is not true in general.)

*If $\mathcal{U}$ is a $\kappa$-closed ultrafilter on an infinite cardinal $\kappa$, we can form the "ultrapower of the universe" $\prod_\kappa V/\mathcal{U}$. It takes a bit of thought to define this properly, but it can be done.

*Since $\mathcal{U}$ is countably closed, the ultrapower is well-founded (think by contrast about an ultrapower of a well-founded model of ZFC by a non-countably-closed ultrafilter), and it's clear from the definition (once written out carefully) that it's set-like, so it is isomorphic to an inner model $M$ of $V$.

*Let $j:V\rightarrow M$ be the embedding induced by this construction (the composition of the Mostowski collapse and the ultrapower embedding). This map $j$ is an elementary embedding, so $M\models j(\kappa)$ is the least measurable cardinal.

*But it's not hard to show that $j(\kappa)>\kappa$. (Think, in the ultrapower, about the equivalence class of the identity function $id:\kappa\rightarrow\kappa$ - this corresponds to an element of $j(\kappa)$, but is above each constant sequence $x\mapsto \alpha$ for each $\alpha\in\kappa$.). This means that $M\not=V$, since $M$ and $V$ disagree about the least measurable.

*So if there is a measurable cardinal, then there is a proper inner model of ZFC. But $L$ is the minimal inner model of ZFC, and so measurable cardinals can't exist in $L$.
The motto here is:

Big large cardinals - measurables and above - imply that the universe is "wide" (specifically, that there are lots of nontrivial inner models). This contradicts V=L since $L$ is "thin."

By contrast, an inaccessible cardinal is "all about height" in a precise sense (namely, inaccessibility is a $\Pi_1$ property). $L$ is no shorter than the whole universe! The naive idea that "large cardinals = high infinities" misses the real deep features of strong large cardinal axioms.
