Estimate with spectral radius and Frobenius inner product

Question

Is the following inequality true for a matrix $C \in \mathbb{R}^{n \times n}$? $$ r(C) \leq \sqrt{|C^{T}:C|}$$ $r(C)$ denotes the spectral radius of $C$ and $X:Y$ denotes the Frobenius-Innerproduct of $X$ and $Y$. It is known that $r(C) \leq \sqrt{C:C}=|C|_F $ and $\sqrt{|C^{T}:C|} \leq |C|_F, $ so that might be true. (If it matters: $C$ is of the form $C=AB$ where A is symmetric and positive-definite and B just symmetric. Can we assume in this case $C^{T}:C \geq 0$?)

Answer 1

In general, no, but in your specific case, yes.

For a counterexample, consider $C=\pmatrix{0&-1\\ 1&0}\oplus I_2$.

In your case where $C=AB$ for some symmetric positive definite matrix $A$ and symmetric matrix $B$, all (complex) eigenvalues of $C$ are real. Therefore $\operatorname{tr}(C^2)=\sum_i\lambda_i(C)^2\ge|\lambda|_\max(C)^2$.