How do you find the Triangle Inequality from an Inertia Matrix?

Question

If you have an inertia matrix of the form
$$\begin{pmatrix} I_{xx} & I_{xy} & I_{xz} \\ I_{yx} & I_{yy} & I_{yz} \\ I_{zx} & I_{zy} & I_{zz} \end{pmatrix}=I$$
If the matrix is symmetric ($I_{xy} = I_{yx}$, $I_{xz} = I_{zx}$, $I_{yz} = I_{zy}$) and positive definite (all its eigenvalues are positive)
How can you show the Triangle Inequalities $I_{xx} + I_{yy} \geq I_{zz}$, $I_{xx} + I_{zz} \geq I_{yy}$, and $I_{yy} + I_{zz} \geq I_{yy}$?

I don't seem to understand how these relationships work or even how one would start a derivation to show them.

Answer 1

The three inequalities follow from

$$I_{xx}=\sum_k m_k(y_k^2+z_k^2),$$

$$I_{yy}=\sum_k m_k(x_k^2+z_k^2),$$

and

$$I_{zz}=\sum_k m_k(x_k^2+y_k^2).$$

They do not follow from the symmetry and positive-definiteness conditions that you mention. For example, the diagonal matrix with diagonal elements 1, 3, and 5 is symmetric and has positive-definite eigenvalues but does not obey one of the triangle inequalities.