It is well known that rigid body inertia tensors are 3 by 3 positive semidefinite matrices, which is the same as saying that their eigenvalues are all non-negative.
A little less known is the fact that those eigenvalues also satisfy the triangle inequalities, which means that the sum of any two eigenvalues is always greater or equal than the remaining one. This arises from the definition of the tensor itself which involves integrals of an always non-negative distribution of mass. Per https://physics.stackexchange.com/a/48273/116038 :
In other words, if a semi-positive definite symmetric real $3×3$ matrix with non-negative eigenvalues […] does not satisfy the triangle inequality (1), it doesn't represent a physically possible distribution of mass.
When estimating inertia tensors using, for example, regression techniques, one can obtain matrices that do not represent physically possible rigid bodies (if we do not constrain the regression solution to physically consistent values).
Using a non-positive semidefinite $3×3$ symmetric matrix in places where an inertia tensor is expected — e.g., in some rigid-body formulations, in simulation, in control schemes, etc. — would entail major problems. For example, a non-positive semidefinite matrix used as inertia tensor would give rise to negative kinetic energies,
$ E_k = \frac{1}{2} \omega^\top I \omega $.
Now, my question is:
What problems (in formulations, simulation, control, etc.), if any, can arise if we use as inertia tensor some matrix that, while satisfying the positive semidefinite condition, do not satisfy the triangle inequality conditions on the eigenvalues?
Update: I've checked that "inertia tensors" that are positive semidefinite but do not satisfy the triangle inequalities still verify the energy conservation law with respect to kinetic energy and work.
Best Answer
A rigid body's principal moments of inertia are obtained from these equations :
$$I_1=\int_V\,(x_2^2+x_3^2)\,\rho\,dV$$ $$I_2=\int_V\,(x_3^2+x_1^2)\,\rho\,dV$$ $$I_3=\int_V\,(x_1^2+x_2^2)\,\rho\,dV$$
where $x=x_1~,y=x_2~,z=x_3$ and the inertia tensor is:
$$I= \left[ \begin {array}{ccc} I_{{1}}&0&0\\ 0&I_{{2}}&0 \\ 0&0&I_{{3}}\end {array} \right] $$
with $$i_\alpha=\int_V\,x_\alpha^2\,\rho\,dV> 0~,\alpha=1,2,3$$ thus: $$I_1=i_2+i_3$$ $$I_2=i_3+i_1$$ $$I_3=i_1+i_2$$
and $$I_1+I_2=i_1+i_2+2\,i_3=I_3+2\,i_3 > I_3$$ $$I_2+I_3=2\,i_1+i_2+i_3=I_1+2\,i_1 > I_1$$ $$I_3+I_1=i_1+2\,i_2+i_3=I_2+2\,i_3 > I_2$$
thus the triangle inequality is a physical feature of a rigid body inertia tensor. If the rigid body is symmetric then the symmetry axes are principal axes and the principal moment of inertia must obey the triangle inequality, otherwise you don't describe the rigid body that you want to describe.