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

I 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. All these questions are part of a more general open problem (you already know about this I presume), which asks to prove \begin{equation*} f(s(S))f(s(T))\quad \prec_w\quad f(s(ST)), \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 2The 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 tracenorm version of the conjecture holds true (Thanks for Betrand for essentially pointing this out). 

