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!!

  • $\begingroup$ Try setting all $x_i$ equal to $1$. $\endgroup$ – Yemon Choi Jun 1 '12 at 7:12
  • $\begingroup$ Yemon, are you sure? I would expect the answer here not to be in closed form for n large enough. $\endgroup$ – Gjergji Zaimi Jun 1 '12 at 7:50
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    $\begingroup$ Gjergji, I understood the question as asking for the smallest $c$ independent of $n$ $\endgroup$ – Yemon Choi Jun 1 '12 at 8:07
  • $\begingroup$ Compute the case where the numbers are in geometric progression? $\endgroup$ – Charles Matthews Jun 1 '12 at 8:35
  • $\begingroup$ We can show that for all natural $n\leq9$ we have $c=1-\frac{1}{n}.$ $\endgroup$ – Michael Rozenberg Jan 31 '19 at 8:23

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.

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    $\begingroup$ $n^{1/3}$ is the right order of magnitude: take $1,n^{-1/3},\dots,n^{-1/3}$. One can fight for the exact constant, of course, but I doubt it makes any difference to the OP whether it is $2^{-5/3}n^{1/3}$ as in Yulia's proof, or $2^{-2}n^{1/3}$ as in my simple-minded example. $\endgroup$ – fedja Jun 1 '12 at 12:10
  • $\begingroup$ Thanks for your help, this is really a good approach!! If it's true I think I have to try another way to attack my problem :( $\endgroup$ – xwangae Jun 2 '12 at 1:59

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:


  • $\begingroup$ These Kantorovich-type inequalities are sort of a different flavor, and arise when you have control over the upper and lower bounds on the variables. $\endgroup$ – Gjergji Zaimi Jun 1 '12 at 10:57
  • $\begingroup$ Well, we don't know (yet) the context in which the problem arised. Maybe the OP has such control. I hope he will tell us. $\endgroup$ – Felix Goldberg Jun 1 '12 at 11:38
  • $\begingroup$ Thanks for your help, here $x_i$ is the frequency of number $i$, so it ranges from 0 to m, where m is given. I'm reading it but some of them requires $x_i \neq 0$ and this is not satisfied in this context. $\endgroup$ – xwangae Jun 2 '12 at 1:54
  • $\begingroup$ Then you can apply the inequalities to $y_{i}=x_{i}+1$. $\endgroup$ – Felix Goldberg Jun 4 '12 at 21:12

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.

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    $\begingroup$ I believe you consider the notation $\sum_{sym}$ standard, by many of us will have a tough time trying to guess what it means, so you'd better have it explained. $\endgroup$ – Seva Jan 31 '19 at 9:06
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    $\begingroup$ @Seva I added something. See now. Wow!!! Seva, you don't remember me? $\endgroup$ – Michael Rozenberg Jan 31 '19 at 9:22

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