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.

Not sure if this is a good approach, but:
Let $f(t) = \sqrt{t+2}-\sqrt{t+1}$. Observe $f$ is strictly decreasing. Then subtract $\sqrt{x+1}+\sqrt{y+1}+\sqrt{z+1}$ from each part of the equation to get: $$\begin{align}
&f(y)+f(z)+f(z+1)\\
=&f(z)+f(x)+f(x+1)\\
=&f(x)+f(y)+f(y+1)
\end{align}$$
This reduces to two variables in each part of the equation. To simplify further, I will subtract each term from $f(x)+f(y)+f(z)$: $$\begin{align}
&f(x)-f(z+1)\\
=&f(y)-f(x+1)\\
=&f(z)-f(y+1)
\end{align}$$
Suppose without loss of generality that $x$ is the largest of $(x, y, z)$. Because $f$ is strictly decreasing, $f(x) \leq f(y)$ and $f(x + 1) \leq f(z + 1)$, so $f(y) - f(x + 1) \geq f(x) - f(x + 1) \geq f(x) - f(z + 1)$ with equality holding only when $x=y=z$. But, equality holds by the equation system above, so we have our result.
