# Let $f(x,y,z):= \sqrt{x+1}+\sqrt{y+2}+\sqrt{z+3}$, such that $f(x,y,z)=f(y,z,x)=f(z,x,y)$ holds. Prove that: $x=y=z$. [duplicate]

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}

or

$$y_{j}z_i$$

and

$$z_{j}>z_i\implies y_{j}

or

$$z_{j}y_i$$

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}

and

$$z_{j}>z_i\implies y_{j}>y_i$$

or

$$z_{j} 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?

• @JohnDouma Because he is not interested in confirming the correctness of my solution. Just sent the question. Commented Aug 26, 2022 at 3:00
• Your equality after "using conjugate" (not sure what that means in this context) isn't true. Just let $x=y=z=-1$. Commented Aug 26, 2022 at 3:39
• The problem itself appears in a recent question. However, since this question is asking for a verification of a solution, it wouldn't seem to count as a duplicate.
– Blue
Commented Aug 26, 2022 at 6:47
• @nonstudent: I think the idea of the proof is quite clever, although the presentation is a bit over-complicated. I might suggest something like this: Consider $x$ fixed, $y\neq x$, and $z\neq x$. From the first equality, gather $x$-, $y$-, $z$-terms together and re-write via conjugation to get [your first eqn, no subscripts]. Since the left-hand side is fixed and the right-hand side is a sum of positive values, "larger/smaller" values of $y$ (relative to $x$) must correspond to "smaller/larger" values of $z$; ie, $y>x \iff z<x$ and $y<x\iff z>x$. On the other hand, ... etc, etc, etc.
– Blue
Commented Aug 26, 2022 at 8:33
• @Blue I learned the some words "...fixed, corresponding", and moreover, I saw that it was possible to do this more quickly with different mathematical steps. Surely you took some time to look at the solution. Please accept my thanks for that humbly. I wish you health and well-being. Commented Aug 26, 2022 at 11:02