# 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, etc., but none of the directions work. Please help me solve this problem.

• Have you tried cancelling out a common term in one of the equalities? For example get rid of the last term in the second equality to get a relation on z and y, then try to prove z=y. Aug 13, 2022 at 10:13
• I don't quite know what you mean, but usually if we want to get rid of $x$ we need a relationship between $x$ and $y,z$. For example$x=y+1$ then we can eliminate $x$ ? Aug 13, 2022 at 10:16
• Cancel $\sqrt{x+3}$ to get $\sqrt{y+1}+\sqrt{z+2}=\sqrt{z+1}+\sqrt{y+2}$ Aug 13, 2022 at 10:18
• I edited, my typo, I'm sorry! Aug 13, 2022 at 10:20
• Are there any restrictions on $x,y,z$? For example, if $x=-2$, then what about $\sqrt{x+1}$? Also, there are two square roots. Do you want only the positive root? Aug 16, 2022 at 20:50

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.
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 $$x = y = z$$.