# Ratio sum comparison on operators

It is known by the Lidskii inequality, that $\sum_{i=1}^n \left|s_i(S)-s_i(T)\right|\le\sum_{i=1}^n s_i(S+T)$, where $s_i(S)$ is the $i$-th singular value of $S$. How would one prove that $$\sum_{i=1}^n \frac{ \left|s_i(S)-s_i(T)\right| }{ \left(s_i(S)+1\right) \left(s_i(T)+1\right) } \le \sum_{i=1}^n \frac{ s_i(S+T) }{ s_i(S+T)+1 }$$ for symmetric matrices $S$ and $T$ if the smallest eigenvalue of $S$ is less than $0$, or is there any counter example?

-
I took the liberty of cleaning up the LaTeX slightly –  Yemon Choi Feb 8 '12 at 8:45
Have to think a bit; an argument along the lines of my previous (wrong) answer, does apply if $S$ and $T$ are positive definite. In the general case, a different approach is needed. –  Suvrit Feb 9 '12 at 2:37
btw do you happen to have any intersection with: mathoverflow.net/questions/84425/on-an-eigenvalue-inequality/… –  Suvrit Feb 9 '12 at 3:46
@Ktb: it is not clear to me that your recent edits are making the question clearer. Moreover, it is slightly bad form to make these changes without leaving the old version up for reference, because then comments or answers made to the older versions are then left without their original, proper context –  Yemon Choi Feb 9 '12 at 20:34
@YemonChoi, Suvrit says the relevant claim has been proved by Audenaert, see new edit to his answer, below. –  Will Jagy yesterday

EDIT (20.10.2014). This claim has now finally been proved (by the person that I thought who would prove it, so nice!!). Amazingly, the solution is simple, once the correct tool, (in this case the Thompson-Freede majorization inequalities) has been invoked.

Please see the nice preprint: "A generalisation of Mirsky’s singular value inequalities", K. M. R. Audenaert, (Oct 2014), for the proof.

I'm leaving the information / material down there to maintain historical context / continuity

I am noting down the following as an answer, also to highlight the real question here.

Your alleged inequality is another example of an inequality of the type discussed in (your??) previous questions: On an eigenvalue inequality and another eigenvalue inequality question.

\begin{equation*} |f(s(S))-f(s(T))|\quad \prec_w\quad f(s(S-T)), \end{equation*} where $f$ is a nonnegative concave function on nonnegative reals such that $f(0)=0$, and $s(\cdot)$ is the singular value map. We apply $f$ to $s(S)$ componentwise.

The previous eigenvalue questions linked to above use $f(x) = x^{1/3}$, the above question essentially reduces to $f(x)=x/(1+x)$.

This problem has been known to be an open problem for a more than a year now, though formally it has been acknowledged as an open problem only very recently (see Audenaert and Kittaneh, Conjecture 7). I mention in passing that Audenaert and Kittaneh attribute this conjecture to Miao, though I came upon it myself a few months ago.

The progress so far is mentioned in my answer to (your?) previous eigenvalue questions. The general case seems much trickier.

## Update 2

The previous claim in the arXiv preprint (originally from 5th sep, 2012) was actually wrong, so this question is as of now fully open, with not even the minor progress that was claimed.

Previous update: It seems that yesterday (5th Sep, 2012), an arXiv preprint was released that seems to have made substantial progress on this problem. In particular, they [claimed to] show the trace-norm version of the conjecture holds true (Thanks for Betrand for essentially pointing this out).

-
My own thanks, Suvrit, for elucidating what these questions have been getting at. I can imagine reasons why people have been reluctant to provide background, but I must say I wish they had been more upfront. –  Yemon Choi Feb 9 '12 at 4:44
You're welcome Yemon. I too wish that people were more direct. –  Suvrit Feb 9 '12 at 13:49
So is either of your authors Ktb? –  Will Jagy Sep 6 '12 at 19:00
@Will: I have no idea; though probably the answer to your question is "no".... –  Suvrit Sep 6 '12 at 21:32
Man-Chung Yue and Anthony Man-Cho So's solution to the majorization conjecture is far from complete. –  Betrand Sep 6 '12 at 21:33