I'm have some difficulties in bounding the following inequality:


I want to find a c as small as possible s.t.
$$\sum\_{i=1}^nx\_i^4\sum\_{i=1}^nx\_i^2 -\sum\_{i=1}^nx\_i^6 \leq c\left(\sum\_{i=1}^nx\_i^3\right)^2$$


where $x\_i$ are all non-negative


I know from the cauchy-inequality that 


$$\sum\_{i=1}^nx\_i^4\sum\_{i=1}^nx\_i^2 \geq \left(\sum\_{i=1}^nx\_i^3\right)^2$$


But I think it useless in my question..


And more generally for some k and l, find out a small c s.t.
$$\sum\_{i=1}^nx\_i^{2k-l}\sum\_{i=1}^nx\_i^l -\sum\_{i=1}^nx\_i^{2k} \leq c\left(\sum\_{i=1}^nx\_i^k\right)^2$$


Anyone help with out?
Thanks!!



On the inequality $\sum_{i=1}^nx_i^4\sum_{i=1}^nx_i^2 -\sum_{i=1}^nx_i^6 \leq c\left(\sum_{i=1}^nx_i^3\right)^2$
I will try to give an estimate. Represent your inequality as:
$$ \sum\_{i\lt k} (x\_i^4x\_k^2+x\_i^2x\_k^4) \le c(\sum\_i x\_i^6 + \sum\_{i\lt k} x\_i^3x\_k^3). $$
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.


