I will try to give an estimate. Represent your inequality as(sum without indices is over $i\lt k$, I cannot make TeX render it):$$\sum sum_{i\lt k} (x_i^4x_k^2+x_i^2x_k^4) \le c(\sum_i x_i^6 + \sum sum_{i\lt k} x_i^3x_k^3).$$There are $n(n-1)/2$ pairs of $i\lt k$, and every $i$ comes in $n-1$ pairs. Distributing this into pairs, we have:\sum_{i\lt k} \frac{c}{n-1}(x_i^6 +x_k^6) + 2cx_i^3x_k^3 - x_i^2x_k^4-x_k^2x_i^4 \ge0.Denote $a=(n-1)/c$ and consider one single pair with $x_i=x$, $x_k=y$:x^6 + y^6 + 2(n-1)x^3y^3 - ax^2y^4-ax^4y^2\ge0.All monomials are uniform (or what is the term?), so we can assume that $y=1$:x^6 + 2(n-1)x^3 - ax^2-ax^4 +1 = (x^2+1) (x^4-(a+1)x^2+1) +2(n-1)x^3\ge0.The biquadratic polynomial $(x^4-(a+1)x^2+1)$ has minimum at $x_0^2=(a+1)/2$, and this minimum equals $1-(a+1)^2/4 = 1-x_0^4$. If $x_0\le1$, i.e. $a=1$, then this is nonnegative and the whole expression is nonnegative. Thus, we already have an estimate: $a_{max}\ge 1$, $c_{min}\le n-1$.
Take now the term with $x^3$ into consideration. Still at the minimum point $x_0$, we have:$$(x_0^2+1)(1-x_0^4)+2(n-1)x_0^3\ge0.Of course we are interested in x_0\ge1 and n\ge3. One estimate I can guess is to put 2(n-1)=\alpha x_0^3, then we want that:$$ (\alpha-1)x_0^6-x_0^4+x_0^2+1\le0,what is of course true for all $x_0\ge1$ if $\alpha=1$, i.e. $x_0=(2(n-1))^{1/3}$. This gives an estimate on $c$ as something like $2^{-5/3}n^{1/3}$... By my methods one scarcely gets much better.
I will try to give an estimate. Represent your inequality as (sum without indices is over $i\lt k$, I cannot make TeX render it): $$\sum (x_i^4x_k^2+x_i^2x_k^4) \le c(\sum_i x_i^6 + \sum x_i^3x_k^3).$$