I have posted this question on Stackexchange but it has received no answer so far. It is a challenging generalization of several difficult inequalities, where none of the usual methods used in inequalities seems to work. I would like to know if someone has a deeper insight into what is happening here.
Let $n\in\mathbb N$, $\lambda> 1$ and put $k:=\lambda n$ and $c:=\frac{\lambda-\frac 1n}{\lambda-1}=\frac{k-1}{k-n}$. For reals $a_1,...,a_n\ge 0,$ define $ r:=\prod_{i=1}^n a_i^c$. Then prove or disprove $$ {\sum\limits_{i=1}^n\displaystyle\dfrac {1-a_i }{1 - a_i^{\lambda n}}\geqslant \frac {1-r}{1 -r^{\lambda}}}.$$
For $\lambda\in\mathbb N$, we can write this inequality in a "purely reciprocal" form as$$ {\sum\limits_{i=1}^n\displaystyle\dfrac {1}{1 + a_i + \cdots + a_i^{k-1}}\geqslant \frac {1}{1 + r+ \cdots +r^{\lambda-1}}.}$$ The limit case $\lambda\to 1$ yields
$\sum\limits_{i=1}^n\displaystyle\dfrac {1}{1 + a_i + \cdots + a_i^{n-1}}\geqslant 1 \qquad\qquad\ \ \text{ if}\ \ \prod a_i\leqslant 1$
$\sum\limits_{i=1}^n\displaystyle{\dfrac {1}{1 + a_i + \cdots + a_i^{n-1}}} \geqslant \prod a_i^\frac{1-n}n \qquad\text{ if}\ \ \prod a_i\geqslant 1$
The most interesting case of this is of course $\prod a_i=1$. I have proven it for this case for all $n\in\mathbb N$ by calculus, generalizing the (corrected) idea given here.
Other special cases that have given rise to my general conjecture above:
The case $n=2,k=3$ is equivalent to $ {\dfrac{1}{1+a+a^{2}}+\dfrac{1}{1+b+b^{2}}+\dfrac{1}{1+c+c^{2}}\ge 1 }$, i.e. the case $n=3,\lambda\to1$ with condition $abc=1$.
The case $n=2,k=4$, i.e. $ { \dfrac{1}{1+a^2+a^4+a^6}+\dfrac{1}{1+b^2+b^4+b^6}\ge\dfrac{1}{1+a^3b^3} }$, is Vasc's "gm_33 lemma" and is already difficult to prove. There are several proofs of this around, all ad-hoc and rather technical. See
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=18857 http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1701829#p1701829The case $n=2,k=6$ has been suggested in one of those threads as a conjecture.
The case $n=3,k=6$ could be called "gm_555 Lemma" and has been suggested by myself. This conjecture says: for $a,b,c\geqslant 0$, $ {\sum\dfrac{1}{1+a^3+a^6+a^9+a^{12}+a^{15}}\ge\dfrac{1}{1+a^5b^5c^5}.}$
One of the striking facts of these inequalities is a certain 'asymmetry' w.r.t. the change $a_i\mapsto\dfrac1{a_i}$.
I think that it should be possible, at least for $k\in\mathbb N$, to find proofs similar to the case $n=2,k=4$, but certainly not without the help of a computer. Any ideas on how to tackle this?
