For the sake of simplicity let fist suppose one has a planar body. We chose a Cartesian system such that axes `Ox` and `Oy` are in the plane of the figure and axis `Oz` is perpendicular to the plane of figure.
For the moment of inertia about axis `Oz` we can write
`I_z = oint_S(r^2*dm) = oint_S(x^2+y^2)*dm =int(x^2*dm)+int(y^2*dm) =I_x+I_y`
Therefore for a plane figure the momentum of inertia about an axis perpendicular to the plane of the figure is the biggest.
Now, again for the sake of simplicity suppose the figure is linear along `Ox` .
Its moment of inertia about `Ox` and `Oy` axes are
`I_y = int(x^2*dm) >0`
`I_x =int(y^2*dm) =int(y^2*0) =0`
Thus for a linear figure the moment of inertia is the biggest for an axis perpendicular to the line that represents the figure.
Each 3-dimensional body has some planar predominant distribution and in that plane some linear predominant distribution. (For example for the human body the predominant plane is that containing both legs and arms and the predominant direction is upwards for a standing person).
Answer: The moment of inertia of a body is the least along the direction where the mass is distributed predominant.
First you need to understand the definition of moment of inertia of a body
about say "a-axis":
I(a-axis) = Integral [ rho * r^2 dv ] integrated over all the infinitysimal volume
elements "dv" of the body which are "r" distance apart from the
a-axis. (density = rho, assuming uniformbody)
You can easily understand that minimum moment of inertia means choosing the axis "a-axis" here such that the integral is minimum, and the integral will be minimum when "r^2" will be minimum, so cleverly if we can estimate for which axis the average of "r^2" will be least then we can be sure that this particular axis for which average of r^2 or the "root mean square" is least then the moment of inertia will also be least.
A Thin Rod: very easily you can see that average r^2 will be least if you take the axis to be passing through the center of the rod and along the length of the rod then M.I will be least, because maximum r^2 in this case will be radius^2 of the rod and as the rod is thin so radius is very small. Suppose one takes the axis to be perpendicular to the rod and at the edge of the rod, clearly if the length is "L" then r^2 can go up to "L^2" which is much bigger than radius^2 of the rod.