This is a not a complete answer, but it may be a beginning of one.
Rewrite the inequality in question as
\begin{equation*}
\text{lhs}:=x^{2/3}\Big(\sum_1^4\frac1{x_i}\Big)^{2/3}+y^{2/3}\Big(\sum_1^4\frac1{y_i}\Big)^{2/3}
\le r^{2/3}\Big(\sum_1^4\frac1{r_i}\Big)^{2/3}=:\text{rhs},
\end{equation*}
where
\begin{equation*}
x_i:=\sqrt{u_i},\quad y_i:=\sqrt{v_i},\quad r_i:=\sqrt{u_i+v_i}=\sqrt{x_i^2+y_i^2},
\end{equation*}
\begin{equation*}
x:=\prod_1^4 x_i,\quad y:=\prod_1^4 y_i,\quad r:=\prod_1^4 r_i,
\end{equation*}
$x_i>0$, $y_i>0$.
Fix any positive values of the $r_i$'s and maximize lhs over all positive $x_i$'s and $y_i$'s such that $\sqrt{x_i^2+y_i^2}=r_i$ for all $i$. To do so, take any $j\in\{1,\dots,4\}$ and write
\begin{equation*}
x_j=r_j\cos t,\quad y_j=r_j\sin t
\end{equation*}\begin{equation*}
f(u+v)\le f(u)+f(v)
\end{equation*}
for $t\in(0,\pi/2)$. Then
\begin{equation*}
\frac32\,\frac{d\,\text{lhs}}{dt}=
y^{2/3}\Big(\sum_1^4\frac1{y_i}\Big)^{-1/3}\Big(\sum_1^4\frac1{y_i}-\frac1{y_j}\Big)\frac{x_j}{y_j}
-x^{2/3}\Big(\sum_1^4\frac1{x_i}\Big)^{-1/3}\Big(\sum_1^4\frac1{x_i}-\frac1{x_j}\Big)\frac{y_j}{x_j}.
\end{equation*}
For the lhs to be maximized, the latter expression must equal $0$, for all$u,v$ in $j$$\mathbb R_+^4$, which can be rewritten as
\begin{equation}
\frac{f(a_j)}{(\prod_1^4 a_i)^{2/3}}=\frac{f(b_j)}{(\prod_1^4 b_i)^{2/3}}, \tag{1}
\end{equation}
where $f(u):=(1-u)u^2$,
\begin{equation*}
a_j:=\frac{1/x_j}{\sum_1^4 1/x_i},\quad b_j:=\frac{1/x_j}{\sum_1^4 1/y_i},
\end{equation*}\begin{equation*}
f(u):=-\left(\left(\frac{1}{\sqrt{u_1}}+\frac{1}{\sqrt{u_2}}+\frac{1}{\sqrt{u_3}}
+\frac{1}{\sqrt{u_4}}\right) \sqrt{u_1 u_2 u_3 u_4}\right)^{2/3}.
\end{equation*}
soNote that the $a_j$ andfunction $b_j$'s are$f$ is positive numbers with
\begin{equation}
\sum_1^4 a_i=\sum_1^4 b_i=1. \tag{2}
\end{equation}
If the expressionhomogeneous: $(1-u)u^2$$f(tu)=tf(u)$ for $f(u)$ had not contained the factor $1-u$, we could have concluded that without loss of generality the $a_j$'s and $b_j$'s must be proportional to each other, whence the $x_j$'s and $y_j$'s must be proportional to each other$t\ge0$. So, whence the $u_j$'s and $v_j$'s must be proportional to each other, whence the desired inequality would have followed$f(u+v)=2f(\frac{u+v}2)$.
Of course, the situationIt remains to notice that we actually have here$f$ is much more complicatedconvex. It appears that the ratio lhs/rhs has multiple maximum pointsIndeed, even aside from the mentioned "proportional" case.
So far, we have 6 equations (four eqs. indeterminant of the Hessian matrix
\begin{equation}
M:=(\frac{\partial^2 f}{\partial u_i\partial u_j})_{i,j=1}^4
\end{equation}
is (1)$0$, and two eqs. in (2)) with 8 unknowns $a_i,b_i$. This (unfortunately) leavesthe principal minors of (at least) two free parameters. Of course$M$ are manifestly positive, one can try other variations (say of lhs leaving rhs invariant); but those additional variations seem even harder to graspafter some algebraic simplifications.
