How to explain visually and in a mathematically simple manner why the moment of inertia of a system is minimum at its center of mass? I've been solving different problems related with finding the moment of inertia in a set of different particles, and ojects of known rotational inertias. Let's say spheres, cylinders, rings, rods and so on.
To put it in perspective, let's assume that we have a cube with four point masses in each corner and I want to find the moment of inertia of the whole system when the moment is at minimum.
I'm assuming that when the degree of rotation (i'm not sure if the term is correct, please indicate me if this is wrong) is the least then the contribution of the rotational inertia is not affected by any additional radius as when there is some offset from the center of mass of the object.
Typically when I want to know what will be the rotational inertia of a system I would use the principle of superposition of rotational inertia for the system and then Steiner's theorem to find the moment of inertia for an axis paralell to that point.
Hence this formula is expressed as:
$I_{p}=I_{o}+mr^2$
where $I_{p}$ is the rotational inertia of the system about that point and $I_{o}$ the moment of inertia where axis of rotation for that object is in its center. Since there isn't any additional radius to add, the whole expression would be reduced to 
$I_{p}=I_{o}$
But I'm not sure if this is a valid conclusion.
There exists a question which has attended this topic but in a mathematical aspect but it is lacking of a justification in the sense of how to visually explain this fact. 
Therefore, does it exist a video or an animation or an experiment on how can this be proved? I don't know if at this point I'm explaining my major confusion. Can all of this be proved in layman terms?.
The existing mathematical proofs for this are a bit too sophisticated for my current knowledge of mathematics, but does there exist an introductory calculus explanation for this behavior? Can someone help me?
 A: I believe that the Steiner principle can be used to show that axis through the centre of mass has the lowest moment of inertia (across the parallel axes), but it doesn't explain why. First, the principle is not intuitive. Second, to prove it you would use a construction similar to what is needed to show the statement in the first place.
Here is my atempt to give intuition to the proof of the statement.
If we project everything to a plane perpendicular to the axes of interest, we basically want to find such a point $\vec y$, that sum of weighted squares
$$
I=\sum m_i(\vec x_i-\vec y)^2
$$
is minimal. This is a weighted least squares problem.
Before showing this in general case, which can be overwhelming at first, let's consider a particular example of this problem in 1d case for two point masses $m_1$ and $m_2$:
$$
I=m_1(x_1-y)^2 + m_2(x_2-y)^2.
$$
Let's expand the squares and group as powers of $y$:
$$
I = (m_1+m_2)y^2 -2(m_1x_1+m_2x_2)y+(m_1x_1^2+m_2x_2^2).
$$
The plot of function $I(y)$ is a parabola. We can find the position of the parabola tip by completing the square:
$$
I=(m_1+m_2)\left(y^2-2\frac{m_1x_1+m_2x_2}{m_1+m_2}y+\left(\frac{m_1x_1+m_2x_2}{m_1+m_2}\right)^2\right)-\frac{\left(m_1x_1+m_2x_2\right)^2}{m_1+m_2}+(m_1x_1^2+m_2x_2^2) =\\
(m_1+m_2)\color{red}{\left(y-\frac{m_1x_1+m_2x_2}{m_1+m_2}\right)^2}+
\color{blue}{(m_1x_1^2+m_2x_2^2)-\frac{\left(m_1x_1+m_2x_2\right)^2}{m_1+m_2}}
$$
So now we see that $I$ contains two terms: blue one, that doesn't depend on $y$ and red one that depend on $y$ and cannot be negative. So the minimal value can be achieved when the red term is zero, which means that
$$
y=\frac{m_1x_1+m_2x_2}{m_1+m_2}.
$$
The case of 2d and arbitrary number of points doesn't change the procedure, it just makes the formulas scarier. The idea stays the same.
$$
I=\vec x^2\sum m_i - 2\vec x\cdot\left(\sum m_i\vec x_i\right)+\sum m_i\vec x_i^2 =\\
\left(\sum m_i\right)\left(\vec x^2-2\vec x\cdot\frac{\sum m_i\vec x_i}{\sum m_i}+\left(\frac{\sum m_i\vec x_i}{\sum m_i}\right)^2\right)-\frac{\left(\sum m_i\vec x_i\right)^2}{\sum m_i}+\sum m_i\vec x_i^2 = \\
\left(\sum m_i\right)\color{red}{\left(\vec x-\frac{\sum m_i\vec x_i}{\sum m_i}\right)^2}+
\color{blue}{\sum m_i\vec x_i^2-\frac{\left(\sum m_i\vec x_i\right)^2}{\sum m_i}}.
$$
A: The way I would explain it is non-rigorous, but intuitively clear: The moment of inertia $MI$ of a mass $M$ with respect to an axis is closely approximated by $ MI = M R^2$, where R is the perpendicular distance from the axis to the center of mass of $M$.  (This is very nearly true when the mass M occupies a small roughly spherical or cubical volume and R is fairly large; needs to be modified when M is close to the axis, but the result is roughly the same.)
When $R$ -> $0$, $MI$ obviously is minimized.  It doesn't reach zero unless $M$ is infinitely small.  You might be able to improve your intuitive feel for this if you model the $MI$ of two masses separated radially from an axis by different distances, then see how $MI$ changes as $R$ goes from a large distance to zero, and beyond to the other side of the axis.
