A (reverse)-Minkowski type inequality for symmetric sums 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) 
 A: 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 . 
A: 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.

