 Solution

First express inaccessibility by universally excluding witnesses to its failure. Besides $\kappa$ being an ordinal greater than $\omega$, require that for every $\alpha <\kappa$ there is no function from $\alpha$ cofinal in $\kappa$, and no injection from $\kappa$ into $\mathcal{P}(\alpha )$. The first condition gives regular cardinalhood; the second says that $\kappa$ is a strong limit cardinal. Under the axiom of choice these are exactly the conditions for a strongly inaccessible cardinal.

These are Pi-one formulas modulo ZF: the only unbounded quantifiers universally range over possible functions. Being a function of the specified domain, having bounded or cofinal range in $\kappa$, and being an injection whose values are subsets of $\alpha$ can all be written with bounded formulas in set theory, without using a power-set object as an unbounded existential witness. Consequently inaccessibility is downward absolute to a transitive inner model of ZFC.

Now suppose $\kappa$ is Mahlo in $V$. It is inaccessible in the constructible universe $L$, and every $V$-inaccessible ordinal below it is also inaccessible in $L$. Let $C\in L$ be any set that $L$ regards as a club set in $\kappa$. Closure and unboundedness are absolute properties of this fixed set of ordinals, so $C$ is also a club set in $V$. By Mahloness in $V$, it contains an ordinal that is inaccessible in $V$, hence in $L$. Every $L$-club therefore meets the $L$-inaccessible ordinals below $\kappa$. Thus Mahloness is downward absolute to the constructible universe: