Show there exists a constant such that the inequality holds Prove there exists a  constant $c \in (0,1)$  ($c$ should depend on $n$) such that
$$\displaystyle \sum_{i=1}^{n} x_i^3\cdot x_{i+1} \le c \sum_{i=1}^n x_i^4$$
holds for all  real numbers $x_i$ such that $\displaystyle \sum_{i=1}^nx_i=0$ and $x_{n+1}=x_1$.

Holder's inequality with $p=4/3$ and $q=4$ shows that $\sum x_j^3x_{j+1}\le\sum|x_j^3x_{j+1}|\le \sum x_j^4$.
If we have equality then the condition for equality in Holder says that $|x_{j+1}|=c|x_j|$ for some $c>0$. Hence $|x_1|=c^n|x_1|$, so $c=1$. So all the $x_j$ have the same absolute value. Assume not all the $x_j$ vanish. Then none of them vanish; equality in the first inequality shows that  then we must have $x_j^3x_{j+1}> 0$ for all $j$. That is, $x_jx_{j+1}>0$. So all the $x_j$ have the same sign, hence $\sum x_j\ne0$.
So: $\sum x_j^3x_{j+1}<1 $ for every $(x_j)$ with $\sum x_j^4=1$ and $\sum x_j=0$. Compactness gives qed.
