The moment of inertia is minimal when the rotation axis passes through the centerofmass and increases as the rotation axis is moved further from the centerofmass.
I = I_{cm} + Mh^{2}
Here M is the mass, h is the distance from the centerofmass to the parallel axis of rotation, and I_{cm} is the moment of inertia about the center of mass parallel to the current axis.
What is the moment of inertia I_{cm} for a uniform rod of length L and mass M rotating about an axis through the center, perpendicular to the rod?
I = I_{cm} + Mh^{2} → I_{cm} = I  Mh^{2}
Now the moment of inertia about the end of the rod is: I = ML^{2}/3.
The distance from the end of the rod to the center is h = L/2. Therefore:
I_{cm}  = 

ML^{2}    M  ( 

)  ^{2}  = 

ML^{2}   

ML^{2}  = 

ML^{2} 