I'm trying to characterize the behavior of the the quantity: $$A = \frac{\sum \limits_{i = 1}^N x_i^4}{(\sum \limits_{i = 1}^N x_i^2)^2},$$ subject to the constraints that $$ \sum \limits_{i = 1}^N x_i = N, \ x_i \ge 0 \ \forall \ i$$

(I.e., we are working with all integer partitions of size $N$). My hypothesis is that if there exists $x_i \ge N/2$, then $A \ge \frac{x_i^4 + (N-x_i)^4}{(x_i^2 + (N-x_i)^2)^2}$ (which I've verified experimentally for N up to 40). But I'm having trouble proving this hypothesis.

The closest I've come is to make use of the fact that $\sum \limits_{i = 1}^N x_i^2 \le N^2$, which gives an upper bound for the denominator, but the corresponding inequality for fourth powers in the numerator doesn't seem helpful, since we're trying to find a lower bound.

I'd greatly appreciate any advice. Please ask for any clarifications if anything is unclear.

Edit: I've decided this question is better phrased explicitly in terms of partitions. So let $$f(\lambda) = \frac{\sum \limits_{i = 1}^N |\lambda_i|^4}{(\sum \limits_{i = 1}^N |\lambda_i|^2)^2},$$ be a function over partitions $\lambda$ of size $N$ and $|\lambda_i|$ is the size of the $i$-th part. The hypothesis now is that in the domain of all partitions $\lambda'$ in which there exists $\lambda_i$ such that $|\lambda_i| \ge N/2$, $f(\lambda')$ is minimized by the 2-element partition $<\lambda_i, \lambda_j>$ (in which $|\lambda_j| = N - |\lambda_i|$).

  • $\begingroup$ Do you mean "at least one of the $x_i$ is greater or equal to $N/2$" or "all $x_i \ge N/2$"? $\endgroup$ – mvw Apr 23 '15 at 14:44
  • $\begingroup$ The former, so we're not limited to partitions of the form <N/2,N/2>. $\endgroup$ – James Evans Apr 23 '15 at 14:46
  • $\begingroup$ Regarding your upper bound, this seems to be a counter example: $3 + 10 + (-10) = 3$ but $9 + 100 + 100 = 209> 9$. $\endgroup$ – mvw Apr 23 '15 at 14:51
  • $\begingroup$ I should have mentioned that all the $x_i$ must be non-negative. $\endgroup$ – James Evans Apr 23 '15 at 14:52
  • $\begingroup$ Would this question be better suited for MathOverflow? $\endgroup$ – James Evans Apr 26 '15 at 12:55

You can assume that $x_N\ge N/2$ and so you want to prove that $$ \frac{\sum \limits_{i = 1}^N x_i^4}{\left(\sum \limits_{i = 1}^N x_i^2\right)^2}\ge \frac{x_N^4 + (N-x_N)^4}{(x_N^2 + (N-x_N)^2)^2} $$ Set $\lambda:=N-x_N$, and assume $\lambda>0$ (else the inequality is trivially true), and set also $$ x:= \frac{x_N}{\lambda},\quad\text{and}\quad y_i:=\frac{x_i}{\lambda},\quad\text{for $i=1,\dots, N-1$}. $$ The inequality now reads: $$ \frac{x^4+\sum \limits_{i = 1}^{N-1} y_i^4}{\left(x^2+\sum \limits_{i = 1}^{N-1} y_i^2\right)^2} \ge \frac{x^4 + 1}{(x^2 + 1)^2}, $$ and you want to prove this inequality under the conditions $x\ge 1$ (corresponding to $x_N\ge N/2$) and $S_1:=\sum_{i=1}^{N-1}y_i=1$, (corresponding to $\sum_{i=1}^N x_i=N$). Set $S_2:=\sum_{i=1}^{N-1}y_i^2$ and $S_4:=\sum_{i=1}^{N-1}y_i^4$. Then you want to prove that the inequality $$ (S_4+x^4)(x^2+1)^2\ge (x^4+1)(S_2+x^2)^2 $$ holds, or equivalently, that for the function $$ f(x):=S_4 x^4+2 S_4 x^2+S_4+2x^6-2S_2 x^6-2x^2S_2-S_2^2-S_2^2x^4 $$ we have $f(x)\ge 0$ for $x\ge 1$. Below we will prove the following inequality: $$ (MI) \qquad\qquad\qquad 1+2S_4\ge 2 S_2+S_2^2. $$ From this inequality it follows that $f(1)=4S_4+2-4S_2-2S_2^2\ge 0$. Hence it suffices to prove that $f'(x)\ge 0$ for all $x\ge 1$. But $$ f'(x)=4S_4 x^3+4 S_4 x+12 x^5-12 S_2 x^5 -4 S_2 x -4 S_2^2 x^3, $$ and so $f'(x)=2x g(x)$ with $$ g(x):=2 S_4 x^2+2S_4+6 x^4-6 S_2 x^4-2S_2-2S_2^2 x^2 $$ and we have to prove that $g(x)\ge 0$ for $x\ge 1$. From $(MI)$ we obtain $$ 2S_4 x^2-S_2^2 x^2\ge 2x^2 S_2-x^2\quad\text{and}\quad 2S_4-2S_2\ge S_2^2-1, $$ which implies $$ g(x)\ge 2 S_2 x^2-x^2+S_2^2-1+6x^4(1-S_2). $$ Now clearly $S_2\le 1$ and so $6x^4(1-S_2)\ge 6x^2(1-S_2)$. We arrive at $$ g(x)\ge S_2^2+(x^2-1)+4x^2(1-S_2)\ge 0, $$ which concludes the proof.

Finally we prove $$ (MI) \qquad\qquad\qquad 1+2S_4\ge 2 S_2+S_2^2. $$ We will prove a slightly more general result: Take $x_1,\dots,x_n$ with $x_i\ge 0$ and set $$ S_1:=\sum_{i=1}^n x_i,\quad S_2:=\sum_{i=1}^n x_i^2\quad\text{and}\quad S_4:=\sum_{i=1}^n x_i^4. $$ Then we have $$ S_1^4+2S_4\ge 2S_2 S_1^2+S_2^2. $$ This inequality, in the case $S_1=1$, gives $(MI)$.

Let's prove the inequality $ S_1^4+2S_4\ge 2S_2 S_1^2+S_2^2 $:

Adding $S_2^2$ and subtracting $2S_4$ and $2 S_2 S_1^2$ it is equivalent to $$ S_1^4-2S_2 S_1^2+S_2^2\ge S_2^2-2S_4+S_2^2, $$ hence we have to prove $$ (S_1^2-S_2)^2\ge 2(S_2^2-S_4). $$ Now we compute $$ S_1^2-S_2=\left(\sum x_i\right)\left(\sum x_j\right)-\sum x_i^2=\sum_{i\ne j}x_i x_j $$ and $$ S_2^2-S_4=\left(\sum x_i^2\right)\left(\sum x_j^2\right)-\sum x_i^4=\sum_{i\ne j}x_i^2 x_j^2. $$ Subtracting $\sum_{i\ne j}x_i^2 x_j^2$ from both sides of the inequality $$ \left(\sum_{i\ne j}x_i x_j\right)^2\ge 2\sum_{i\ne j}x_i^2 x_j^2, $$ we see that the inequality is equivalent to $$ \sum_{\begin{array}{c} (i,j)\ne (k,l)\\ i\ne j\\ k\ne l\end{array}}x_i x_j x_k x_l\ge \sum_{i\ne j}x_i^2 x_j^2. $$ But we have $$ \sum_{\begin{array}{c} (i,j)\ne (k,l)\\ i\ne j\\ k\ne l\end{array}}x_i x_j x_k x_l= \sum_{\begin{array}{c} (i,j)\ne (k,l)\\ i\ne j\\ k\ne l \\ (i,j)=(l,k)\end{array}}x_i x_j x_k x_l+ \sum_{\begin{array}{c} (i,j)\ne (k,l)\\ i\ne j\\ k\ne l \\ (i,j)\ne (l,k)\end{array}}x_i x_j x_k x_l $$ hence $$ \sum_{\begin{array}{c} (i,j)\ne (k,l)\\ i\ne j\\ k\ne l\end{array}}x_i x_j x_k x_l\ge \sum_{\begin{array}{c} (i,j)\ne (k,l)\\ i\ne j\\ k\ne l \\ (i,j)=(l,k)\end{array}}x_i x_j x_k x_l=\sum_{i\ne j}x_i^2 x_j^2, $$ which concludes the proof of $(MI)$.

  • $\begingroup$ Note that $f(1)\ge 0$ proves the inequality $(MI)$. So any proof of the original hypothesis proves also $(MI)$. This very interesting inequality cannot be proven by Cauchy-Schwartz (at least I'm unable to do it). $\endgroup$ – san May 4 '15 at 18:13

The trick here is to consider the following "switching operation": $$\sigma_{ij}(\lambda_1,\ldots,\lambda_N) = (\lambda_1,\ldots,\lambda_i+1,\ldots,\lambda_j-1,\ldots,\lambda_N).$$ Now the question becomes: what can we say about $f(\sigma_{ij}(\lambda))$?

Assume without loss of generality that $\lambda_j\ge1$, and that all $\lambda_k\ge0$. A short computation shows: $$f(\sigma_{ij}(\lambda)) = \frac{\sum_{k=1}^N\lambda_k^4 + 4(\lambda_i^3-\lambda_j^3) + 6(\lambda_i^2+\lambda_j^2) + 4(\lambda_i-\lambda_j) + 2}{\left(\sum_{\ell=1}^N\lambda_\ell^2+2(\lambda_i - \lambda_j) + 2\right)^2}$$ Now write $\lambda_j = \lambda_i+k$ with $k$ an integer. Then the above expression becomes $$f(\sigma_{ij}(\lambda)) = \frac{\sum_{k=1}^N\lambda_k^4 + 2\lambda_i^2+k(4+2\lambda_i-3\lambda_i^2)+k^2(1-3\lambda_i^2)-k^3}{\left(\sum_{\ell=1}^N\lambda_\ell^2+2(k+1)\right)^2}$$ Now we are left with the tedious work of checking by cases ($k>0$, $k<0$ etc.) when this decreases with respect to $f(\lambda)$. This allows you to minimize the quantity.


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