$\det(A+H)\det(A-H)\le \det(A^2) $ If $H$ is rank $1$, show $\det(A+H)\det(A-H)\le \det(A^2)$.
If $H$ is rank $1$ then we know every column is a multiple of some vector $x$, say $H_{ij}=c_jx_i$. So we can expand the definition of the determinant in the LHS to get a big sum in terms of $A,c,x$. I was trying to simplify this but didn't get anywhere. Am I doing the right thing?
 A: A rank $1$ matrix can be written as $H = uv^{T}$. The matrix determinant lemma states that if $A$ is invertible then $\det(A+uv^{T}) = (1 + v^{T}A^{-1}u)\det (A)$. Thus, for $A$ invertible,
$$\det(A+H)\det(A-H) =  (1 + v^{T}A^{-1}u)(1 - v^{T}A^{-1}u)\det (A)\det(A)$$
$$=(1  - (v^{T}A^{-1}u)^2)\det(A^2) $$
$$\leq \det(A^2)$$
This concludes the proof for $A$ invertible. The proof then follows for all $A$ using the continuity of the determinant function and the density of invertible matrices -- in less-technical language, if you can prove an inequality for invertible matrices and all the functions involved are continuous, the inequality holds for all matrices.
A: If $A$ is invertible, we can divide the inequality by $(\det A)^2$ to obtain an equivalent inequality
$$\det(I+A^{-1}H)\det(I-A^{-1}H) \le 1.$$
Now, the matrix $A^{-1}H$ again has rank $1$ so it has at most one nonzero eigenvalue $\lambda \in \Bbb{R}$ (it is easy to see that it has to be real if $A^{-1}H$ is real).
Then by the spectral mapping theorem we have
$$\det(I+A^{-1}H)\det(I-A^{-1}H) = (1+\lambda)(1-\lambda) = 1-\lambda^2 \le 1.$$
The other case is that $A^{-1}H$ has only zeroes as eigenvalues so $\det(I \pm A^{-1}H) = 1$.
This proves the original inequality for invertible matrices $A$. As Elchanan Solomon suggests, by using density of invertible matrices, we conclude that the inequality holds for all matrices $A$.
