TITLE: A (reverse)-Minkowski type inequality for symmetric sums
QUESTION [10 upvotes]: 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)

REPLY [4 votes]: 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.