Let $(u_1, u_2, u_3, u_4)$ and $(v_1, v_2, v_3, v_4)$ be vectors in $\mathbb R_+^4$. Is the following inequality true?

\begin{align*} \left(\sum_{{[4] \choose 3}} \sqrt{u_i u_j u_k}\right)^{2/3} + \left(\sum_{{[4] \choose 3}} \sqrt{v_i v_j v_k}\right)^{2/3} \leq \left(\sum_{{[4] \choose 3}} \sqrt{(v_i+u_i) (v_j+u_j) (v_k+u_k)}\right)^{2/3} \end{align*} Here $\sum_{{[4] \choose 3}}$ refers to $\sum_{1\leq i < j < k \leq 4}$.

I ran a million Matlab simulations for random vectors and it did not yield any counterexample.

Note: Previously asked on MSE (https://math.stackexchange.com/questions/2647608/a-minkowski-like-inequality-for-symmetric-sums)


Rewrite the inequality in question as \begin{equation*} f(u+v)\le f(u)+f(v) \end{equation*} for $u,v$ in $\mathbb R_+^4$, where \begin{equation*} f(u):=-\left(\left(\frac{1}{\sqrt{u_1}}+\frac{1}{\sqrt{u_2}}+\frac{1}{\sqrt{u_3}} +\frac{1}{\sqrt{u_4}}\right) \sqrt{u_1 u_2 u_3 u_4}\right)^{2/3}. \end{equation*} Note that the function $f$ is positive homogeneous: $f(tu)=tf(u)$ for $t\ge0$. So, $f(u+v)=2f(\frac{u+v}2)$.

It remains to notice that $f$ is convex. Indeed, the determinant of the Hessian matrix \begin{equation} M:=\Big(\frac{\partial^2 f}{\partial u_i\partial u_j}\Big)_{i,j=1}^4 \end{equation} is $0$, and the principal minors of $M$ are manifestly positive, after some algebraic simplifications. (That the determinant of $M$ is $0$ can be shown either by direct calculations or by recalling that $f$ is positive homogeneous and hence $\frac{d^2}{dt^2}\,f(tu)=0$ for $t>0$.)

Dealing with the determinants of the matrices $\Big(\frac{\partial^2 f}{\partial u_i\partial u_j}\Big)_{i,j=1}^k$ for $k=1,2,3$, in view of the positive homogeneity, we may assume without loss of generality that $u_4=1$.

Details of the calculations can be seen in the the Mathematica notebook and/or its pdf image .

  • $\begingroup$ I admire the tour de force solution! I will wait a couple of days before accepting, however, since I would very much like to see a proof that can also generalize to an ${n \choose k}$ setting (the current one being ${4 \choose 3}$.) $\endgroup$
    – VSJ
    Feb 14 '18 at 4:53
  • 1
    $\begingroup$ I think, if you want an answer to a more general or otherwise significantly amended question, it is better to ask it separately. When trying to decide whether to invest time and effort into answering a question, people don't necessarily anticipate that the question may change down the road. $\endgroup$ Feb 14 '18 at 16:10
  • $\begingroup$ I agree; I have no intention of changing this question. $\endgroup$
    – VSJ
    Feb 14 '18 at 17:36

The said claim follows from the following general result on elementary symmetric polynomials, denoted $e_k$ below. $\newcommand{\vx}{\mathbf{x}}\newcommand{\vy}{\mathbf{y}}$

Theorem A (S. 2018). $\,$ Let $p\in (0,1)$ and $x \in \mathbb{R}_+^n$. The map \begin{equation*} \phi_{k,n}(x) := x \mapsto \left[\frac{e_k(x_1^p,\ldots,x_n^p)}{e_{k-1}(x_1^p,\ldots,x_n^p)}\right]^{1/p}, \end{equation*} is concave. Recently, I typed up a proof to this inequality (and a few others). Please see this preprint for a proof.

As a corollary, we obtain a concavity result that implies the OP's conjectured inequality as a special case (using positive homogeneity).

Corollary. Let $p\in (0,1)$ and $\vx \in \mathbb{R}_+^n$. Then, $[e_k(\vx^p)]^{1/pk}$ is concave (we write $\vx^p \equiv (x_1^p,\ldots,x_n^p)$. \begin{align*} [e_k((\vx + \vy)^p)]^{1/pk} &= \left[\frac{e_k((\vx+\vy)^p)}{e_{k-1}((\vx+\vy)^p)}\cdot \frac{e_{k-1}((\vx+\vy)^p)}{e_{k-2}((\vx+\vy)^p)}\cdots \frac{e_1((\vx+\vy)^p)}{e_0((\vx+\vy)^p)} \right]^{1/pk}\\ &= \left[\phi_{k,n}(\vx+\vy)\phi_{k-1,n}(\vx+\vy)\cdots\phi_{1,n}(\vx+\vy)\right]^{1/k}\\ &\ge \left[\left(\phi_{k,n}(\vx)+\phi_{k,n}(\vy)\right)\cdots\left(\phi_{1,n}(\vx)+\phi_{1,n}(\vy)\right)\right]^{1/k}\\ &\ge \prod_{j=1}^k[\phi_{j,n}(\vx)]^{1/k} + \prod_{j=1}^k[\phi_{j,n}(\vy)]^{1/k}\\ &= [e_k(\vx^p)]^{1/pk} + [e_k(\vy^p)]^{1/pk}, \end{align*} where the first inequality follows from Theorem A and the second is just Minkowski.

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    $\begingroup$ A collaborator and I were able to prove the ${n \choose k}$ case of the original question by directly computing the Hessian by hand and showing it is positive semidefinite. Your generalizations and extensions are superb! $\endgroup$
    – VSJ
    Mar 31 '18 at 19:57
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    $\begingroup$ Thanks! Impressive that you could work out a direct proof by hand -- I try to avoid computing Hessians whenever I can! $\endgroup$
    – Suvrit
    Mar 31 '18 at 20:09

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