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.
On
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 $x = y = z$.
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.