Symmetric positive definite matrix inequality

Question

Hi could you help me with the following:

Show that for a symmetric positive definite matrix $B$, $$b_{ij} + b_{jk} + b_{ki} \leqslant b_{ii} + b_{jj} + b_{kk}$$ holds for any $1 \leqslant i,j,k \leqslant n$ with $b_{ij}$ being the entry at $(i,j)$ of matrix $B$.

Thanks a lot.

Answer 1

This reduces to the fact that $b_{ii} + b_{jj} \geq 2b_{ij},$ applied three times: specifically, $b_{ii} + b_{jj} \geq 2b_{ij}$,$b_{ii} + b_{kk} \geq 2b_{ik}$ and $b_{jj} + b_{kk} \geq 2b_{jk}$. Adding the three inequalities, and dividing by $2$ gives the desired inequality. I leave you to verify that $b_{ii} + b_{jj} \geq 2b_{ij}.$

Answer 2

Here is a slightly more general perspective. For any positive diagonal matrix $D$ and real orthogonal matrix $U$, we have $\operatorname{tr}D\ge\operatorname{tr}DU$. Put $U=Q^TPQ$ with a real orthogonal matrix $Q$ and a permutation matrix $P$ and make use of the identity $\operatorname{tr}AB=\operatorname{tr}BA$, we get $\operatorname{tr}QDQ^T\ge\operatorname{tr}QDQ^TP$ and in turn $\operatorname{tr}\tilde{B}\ge\operatorname{tr}\tilde{B}P$ for every positive definite matrix $\tilde{B}$. Now, let $\tilde{B}$ be a 3-by-3 principal submatrix of $B$ and the result follows.

Answer 3

This is just user Geoff's Answer expanded. I saw it after typing up this whole thing.

Consider the $2 \times 2$ principal minors \begin{align} \begin{bmatrix}b_{ii} & b_{ij} \\b_{ji} & b_{jj} \end{bmatrix}~ and ~\begin{bmatrix}b_{ii} & b_{ik} \\b_{ki} & b_{kk} \end{bmatrix}~ and~ \begin{bmatrix}b_{jj} & b_{jk} \\b_{kj} & b_{kk} \end{bmatrix} \end{align} Determinant of all of them should be positive (related to sylvester's criterion). This gives you $b_{ii}b_{jj}\geq b_{ij}^2$ (note $b_{ij}=b_{ji}$) and other two conditions. From $b_{ii}b_{jj}\geq b_{ij}^2$, you get $b_{ii}+b_{jj}\geq 2*b_{ij}$. Similarily for other minors also, add them all together and you have your inequality.