The problem:
Let $x,y,z\in\mathbb R_{\ge -1}$, such that $$\begin{aligned}&\sqrt{x+1}+\sqrt{y+2}+\sqrt{z+3}\\ =&\sqrt{y+1}+\sqrt{z+2}+\sqrt{x+3}\\ =&\sqrt{z+1}+\sqrt{x+2}+\sqrt{y+3}\end{aligned}$$
holds. Prove that:
$$x=y=z$$
My attempts:
Proposition: Let $x=x_i,y=y_i,z=z_i$ be a possible solution combination for some $i\in\mathbb N$. Then for every $x_i$, there exist exactly one possible solution pair: $y=y_i, z=z_i.$
Proof: Suppose that $x=x_i, y=y_{j},z=z_{j}$ is an also possible solution combination for some $j\in\mathbb N$, such that $y_{j}\neq y_i$ or $z_{j}\neq z_i$.
Then using conjugate we have,
$$\frac{2}{\sqrt {x_i+1}+\sqrt {x_i+3}}=\frac{1}{\sqrt {y_i+1}+\sqrt {y_i+2}}+\frac{1}{\sqrt {z_i+2}+\sqrt {z_i+3}}$$
From here we obtain:
$$y_{j}>y_i\implies z_{j}<z_i$$
or
$$y_{j}<y_i\implies z_{j}>z_i$$
and
$$z_{j}>z_i\implies y_{j}<y_i$$
or
$$z_{j}<z_i\implies y_{j}>y_i$$
On the other hand, we have
$$\frac{1}{\sqrt {x_i+1}+\sqrt {x_i+2}}=\frac{2}{\sqrt {z_i+1}+\sqrt {z_i+3}}-\frac{1}{\sqrt {y_i+2}+\sqrt {y_i+3}}$$
From here we also obtain:
$$y_{j}>y_i\implies z_{j}>z_i$$
or
$$y_{j}<y_i\implies z_{j}<z_i$$
and
$$z_{j}>z_i\implies y_{j}>y_i$$
or
$$z_{j}<z_i \implies y_{j}<y_i$$ A contradicton.
Final step: We know that, $x=x,y=x,z=x$ is a possible solution. Then the proposition tells us: for every $x$, the solution $y=x,z=x$ is unique. Thus, we can conclude
$$x=y=z$$
is an only possible solution.
My math background is just I we learned at school age. Therefore, I couldn't be sure of my solution.
Are there missing points in my attempts?