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Let $b$ and $a_i$ be positive real numbers, $i=1,\ldots,N$. Let the geometric mean $G(\{x_i\})=\sqrt[n]{\prod_{i=1}^Nx_i}$.

I know that $G(\{a_i+b\})\geq b+ G(\{a_i\})$, but when does the equality holds? is there a way to write $G(\{a_i+b\})$ as a function of $b$ and $G(\{a_i\})$ (the same way as is the case with the arithmetic mean: $\dfrac{1}{N} \sum (a_i+b)=\dfrac{1}{N}\sum (a_i)+b$)

Toney Shields
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    With $N=2$ you'll find that $G({a_1+b,a_2+b}) = b+ G({a_1,a_2})$ simplifies to $G({a_1,a_2})=\frac{a_1+a_2}2$. I think this then generalises to larger $N$, where the equality holds when the GM equals the AM. For the second question, you can easily find a counterexample using two sets of $a_i$ with the same GM and then add the same $b$ to get two different GMs. – Jaap Scherphuis Aug 16 '22 at 09:45
  • One can take $a_i=c_i^N$, then $G({a_i})=\prod c_i$, inequality turns to $\prod (b+c_i^N) \geq (b+\prod c_i)^N$. WLOG one can take $b=1$, because $b$ can be factored out from all $a_i$. – Ivan Kaznacheyeu Aug 16 '22 at 11:05

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