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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}} $](http://latex.artofproblemsolving.com/c/d/4/cd481672bee1e87d8c97ac623436e0ec75cfa73a.png)
are given such that 
. Prove the inequality 
, let 
denote the smallest possible value of 
where 
are sets such that 
and 
whenever 
. Determine for each positive integer .
![$ \mathbb Z[x]$](http://latex.artofproblemsolving.com/0/3/8/0382040a765290e7da81187825e4f9e75e47afde.png)
with at least one coefficient larger than 1387 such that coefficients of their product is in the set 
.
such that all of its coefficient are in the set
with at least one coefficient larger than 1387 such that coefficients of their product is in the set . such that all of its coefficient are in the set 
. and we can write :
where :
and 
sufficient big both of these polynomials have a coefficient bigger than 
.
where all coefficient of 
are and 
. and we can assume 
: if 
then : 
.
so we must have 
but all of coefficient of are and so we must have 
Thus we can easily prove by lemma 
. Contradiction !
in natural numbers m and n.
