Let $d>1$, and let $\alpha_1,\dots,\alpha_d,\beta_1,\dots,\beta_d$ be positive real numbers. Is there an easy proof of the following claim:
If $\alpha_j-(\Pi_{i=1}^d \alpha_i)^{1/d}=\beta_j-(\Pi_{i=1}^d \beta_i)^{1/d} $ for every $1\le j\le d$ and not all of these differences are zero, then $\alpha_i=\beta_i$ for every $i$.
If we allow all of the differences to be zero, then uniqueness fails: set $\alpha_i=a,\beta_i=b$ for $a \neq b$.
I think I have a proof for this uniqueness, but it is very cumbersome. I am interested in seeing other approaches.
Assume that that condition holds. Then $\beta_i=c+\alpha_i$ for all $i,$ where $c=\prod_{i=1}^d\beta_i^{1/d}-\prod_{i=1}^d\alpha_i^{1/d}.$ If $c=0$ we're done, so assume not. By swapping $\alpha$ and $\beta$ we can assume $c> 0$ without loss of generality.
The $d$-function $L^d$ Hölder inequality applied to the functions $f_i$ defined by $f_i(0)=c^{1/d}$ and $f_i(1)=\alpha_i^{1/d},$ on the domain $\{0,1\}$ with counting measure, gives
$$c+\prod_{i=1}^d \alpha_i^{1/d}\leq \prod_{i=1}^d \|f_i\|_d=\prod_{i=1}^d (c+\alpha_i)^{1/d}=\prod_{i=1}^d \beta_i^{1/d}$$ with strict inequality unless all the $\alpha_i$ are the same. But strict inequality would contradict $c=\prod_{i=1}^d\beta_i^{1/d}-\prod_{i=1}^d\alpha_i^{1/d}.$