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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.

Asaf Shachar
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1 Answers1

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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 $c=\prod_{i=1}^d\beta_i^{1/d}-\prod_{i=1}^d\alpha_i^{1/d}.$

Dap
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  • Wow! How on earth did you think about this use of Holder's inequality? This seems like a very non-trivial approach...Thank you. – Asaf Shachar Dec 11 '18 at 15:02
  • Hölder's inequality for finite sums is a standard tool in "problem solving"/"contest math" style inequalities. Though I had happened to have just read another question where its use was quite surprising to me: https://math.stackexchange.com/questions/3031422/find-all-positive-integers-a-and-b-such-that-1-a8-ba-b-27ab/3031468#3031468 – Dap Dec 11 '18 at 18:55
  • Thank you. I still find this use of Hölder's inequality somewhat miraculous... so I asked another question about it here: https://math.stackexchange.com/questions/3036749/a-simple-proof-for-prod-i-1d-a-i-prod-i-1d-b-i-le-prod-i-1d-a-id. Thank you again for this very nice solution. – Asaf Shachar Dec 12 '18 at 14:29