Prove if $\sqrt{x+1}+\sqrt{y+2}+\sqrt{z+3}=\sqrt{y+1}+\sqrt{z+2}+\sqrt{x+3}=\sqrt{z+1}+\sqrt{x+2}+\sqrt{y+3}$, then $x=y=z$. Let $x$, $y$, $z$ be real numbers satisfying $$
\begin{align}
&\sqrt{x+1}+\sqrt{y+2}+\sqrt{z+3}\\
=&\sqrt{y+1}+\sqrt{z+2}+\sqrt{x+3}\\
=&\sqrt{z+1}+\sqrt{x+2}+\sqrt{y+3}.
\end{align}$$
Prove that $x=y=z$.
I tried assuming $x>y>z$, $x>y=z$,$x<y<z$, etc., but none of the directions work. Please help me solve this problem.

The solution of tehtmi is wonderful, and I have a similar approach.
For each parameter $t \in \{x, y, z\}$ and each $1 \leq i \leq 3$, let $t_i = \sqrt{t + i}$. For example $x_2 = \sqrt{x + 2}$. So we have:
\begin{align*}
&x_1 + y_2 + z_3\\
=\ &y_1 + z_2 + x_3 \label{1}\tag{$*$}\\
=\ &z_1 + x_2 + y_3
\end{align*}
Suppose $x = \min\{x, y, z\}$. Note that the function $f(t) = \sqrt{t + m} - \sqrt{t + n}$ for all $m > n$ is strictly decreasing. Thus
\begin{alignat*}{2}
y_2 - y_1 &\leq x_2 - x_1 &&\implies x_1 + y_2 \leq y_1 + x_2\\
z_3 - z_2 &\leq x_3 - x_2 &&\implies z_3 + x_2 \leq z_2 + x_3\\
&\ &&\stackrel{+}{\implies} x_1 + y_2 + z_3 \leq y_1 + z_2 + x_3
\end{alignat*}
But by \eqref{1} the equal case has occurred, and the equal case occurs only for $x = y = z$.
