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which satisfy 
![$ \frac{a+b+c}{3} \ge \sqrt[10]{\frac{a^3+b^3+c^3}{3}}$](http://latex.artofproblemsolving.com/c/d/a/cdae05ad34239366b0dd3921243522c711b6ae58.png)
method, I can assum 
![$ \frac{2x+\frac{1}{x^2}}{3} \ge \sqrt[10]{ \frac{2x^3+ \frac{1}{x^6}}{3}} $](http://latex.artofproblemsolving.com/c/d/4/cd481672bee1e87d8c97ac623436e0ec75cfa73a.png)
are given such that 
. Prove the inequality 
, let 
denote the smallest possible value of 
where 
are sets such that 
and 
whenever 
. Determine for each positive integer .
and 
such that 
where indices are modulo 
. Let 
. Consider a set 
such that each element of is of the form 
where 
and if we have any 
with 
, then 
should occur in 
elements of . The existence of such a set can be proven using Hall's Marriage Theorem by considering a set to have 
copies of each subset of 
of size 
and then noticing that there is a matching from subsets of size and those of size 
because they can be shown to satisfy Hall's condition and gluing the matchings would give rise to . be any set satisfying the given conditions. Then ![\[|K| = \frac{\sum_{A\in K}\dbinom{n}{|A|}\cdot \sum_{P}^{A\in P \in T} 1}{|T|} = \frac{\sum_{P\in T} \sum _{A\in P\cap K } \dbinom{n}{|A|}}{|T|}\le \frac{\sum_{P\in T} M}{|T|} = M\]](http://latex.artofproblemsolving.com/7/6/c/76c5b3219f67501f15dad96152bd971868f46469.png)
![\[K = \bigcup_{j=0}^{\left\lfloor\frac{n-k}{3}\right\rfloor} \{ S| S\subset\{1, 2, \ldots, n\}, |S|=k+3j\}\]](http://latex.artofproblemsolving.com/6/9/f/69fcfa2250ccef81f4311276a1030526d25f715f.png)
( is even) or 
( is odd) satisfy the requirements of the problem, the number of the subsets in this group is 
or 
one of which equals 
. that the maximum number of subsets is .
is obvious.
, and we will prove for .
subgroups, on of which contains all subsets which do not contain the element 
and the other subgroup contains the rest of the subsets. The first subgroup has at most 
by induction, the second subgroup has at most 
elements (we can consider subsets 
instead of 
if 
, then add to the number of that subsets 
for the subset 
) by induction. This gives the result in case is odd, still for even more tricks must be done.
be the maximum.
.
, from which we can easily finish by induction.
. Note for every 
, at most one of 
is in the family. For the remaining sets (which contain and exclude 
), we can pick at most more sets. Hence our conclusion follows.
subsets into the floor of 
chains of elements and then the last is either a singleton or a chain of , depending on the parity of . The induction step is more or less "clone and combine."
the answer is ? we can get 
subsets whereas you predicted to have at most 
. Here are the subsets, 
. It seems to have at most 
subsets instead. Sorry, in advance if I am wrong somewhere.
in natural numbers m and n.
such that for any two of them like 
if 
then 
.
is the number of elements of the set 
.
_
is set of the elements that are in 
but not in 
the solution of the problem for 
and then continue on knowing what sum of binomial coefficents to take(module 
)
be sets which satisfies the conditions of the problem.
.
and
,
and so 
.
，
.