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'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!!
 A: 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.
A: You can try Ozeki’s inequality, which is one of a number of known "reverse Cauchy-Schwarz inequalities" and seems best adapted to your situation. Have a look here:
http://www.ajmaa.org/RGMIA/papers/v6n4/RCBSInTCN.pdf
A: Let $x_i=1$ for all $i$.
Thus, $c\geq1-\frac{1}{n}.$
We'll prove that our inequality is true for $c=1-\frac{1}{n}$ for all natural $n\leq9.$
Indeed, we need to prove that
$$\frac{n(n-1)}{n!}\sum_{sym}x_1^4x_2^2\leq\left(1-\frac{1}{n}\right)\left(\frac{n}{n!}\sum_{sym}x_1^6+2\cdot\frac{\frac{n(n-1)}{2}}{n!}\sum_{sym}x_1^3x_2^3\right)$$ or
$$\sum_{sym}(x_1^6+(n-1)x_1^3x_2^3-nx_1^4x_2^2)\geq0$$ or
$$\sum_{sym}(x_1^6-nx_1^4x_2^2+2(n-1)x_1^3x_2^3-nx_1^2x_2^4+x_2^6)\geq0.$$
Now, let $x_1=tx_2.$
Thus, it's enough to prove that
$$t^6-nt^4+2(n-1)t^3-nt^2+1\geq0$$ or
$$(t^3-1)^2\geq nt^2(t-1)^2,$$ for which it's enough to prove that
$$(t^2+t+1)^2\geq9t^2,$$ which is true by AM-GM.
I used $\sum\limits_{sym}$ in the following sense. 
For example for four variables $a$, $b$, $c$ and $d$.
We know that $|S_4|=4!=24$.
$a+b+c+d$ it's a sum of $4$ addends, which says $$a+b+c+d=\frac{4}{4!}\sum_{sym}a=\frac{1}{6}\sum_{sym}a$$ or
$$\sum_{sym}a=6(a+b+c+d).$$
Similarly,
$$ab+ac+bc+ad+bd+cd=\frac{1}{4}\sum_{sym}ab$$ or
$$\sum_{sym}ab=4(ab+ac+bc+ad+bd+cd).$$
Also, we have for example
$$\sum_{cyc}a^2(b+c+d)=\frac{1}{2}\sum_{sym}a^2b$$ by the same reasoning.
