Consider the following block matrix
$$
X = \begin{bmatrix} A & C \\ C^\top & B\end{bmatrix},
$$
where $A\in\mathbb{R}^{n\times n}$, $B\in\mathbb{R}^{m\times m}$, and $C\in\mathbb{R}^{n\times m}$.


**To Prove (or disprove)**: If $X$ is positive definite, i.e. $X>0$, then the following trace inequality holds
$$
\left[\mathrm{tr}(CC^\top)\right]^2< \mathrm{tr}(A^2)\mathrm{tr}(B^2).
$$


**Some comments.** Based on Theorem 2.3 of [Horn and Mathias. "Cauchy-Schwarz inequalities associated with positive semidefinite matrices." *Linear Algebra and Its Applications* 142 (1990): 63-82](http://www.sciencedirect.com/science/article/pii/002437959090256C), I think it is possible to prove the previous fact if we replace strict inequalities with non-strict ones.



On a trace condition for positive definite $2\times 2$ block matrices
For any unitarily invariant norm it can be shown that
\begin{equation\*}
 \|X\| = \left\Vert 
 \begin{bmatrix}
 A & C\\
 C^\* & B
 \end{bmatrix}
 \right\Vert \le \|A\| + \|B\|.
\end{equation\*}
Thus, using the squared Frobenius norm on both sides and cancelling, we obtain
\begin{equation\*}
 \|C\|\_F^2 \le \|A\|\_F\|B\|\_F,
\end{equation\*}
as desired.


