I accidentally discovered this equality, in which I can prove numerically using python.
But I need to prove this equality in an algebraic way and I could not get anywhere. The right-hand side is nothing more than an arithmetic mean, while the left-hand side is a modification of the geometric mean. I hope someone can help me at this point.
On
Your expression under limit can be written as $k(a-b) $ where $$a=\sqrt[n]{\prod_{i=1}^{n}\left(1+\frac{x_i}{k}\right)}, b=1$$ and further $a-b=(a^n-b^n) / c$ where $a, b$ tend to $1$ and $$c=a^{n-1}+a^{n-2}b+\dots+ab^{n-2}+b^{n-1}$$ tends to $n$ as $k\to\infty$. And $$a^n-b^n=\prod_{i=1}^{n}\left(1+\frac{x_i}{k}\right)-1=\frac{1}{k}\sum_{i=1}^{n}x_i+o(1/k)$$ and thus $$k(a-b) =\frac{1}{n}\sum_{i=1}^{n}x_i+o(1)$$ and the desired limit as $k\to\infty$ is $\dfrac{1}{n}\sum\limits_{i=1}^{n}x_i$.
On
You could even get an interesting asymptotics considering $$y_n=\sqrt[n]{ \prod_{i=1}^{n}{(x_i+k)}}$$ $$ \log(y_n)=\frac 1n \sum_{i=1}^{n}\log\left({(x_i+k)}\right)=\frac 1n \sum_{i=1}^{n}\left(\log(k)+\log\left(1+\frac {x_i}k\right)\right)$$ Assuming that $x_1<x_2<\cdots < x_n \ll k$, use the Taylor expansion of $\log\left(1+\frac {x_i}k\right)$. Then continuing with Taylor series $$y_n=e^{\log(y_n)}\implies y_n=k+\frac{\sum_{i=1}^{n} x_i}{n}+\frac{\left(\sum_{i=1}^{n} x_i\right)^2-n\sum_{i=1}^{n} x^2_i}{2n^2 k}+O\left(\frac{1}{k^2}\right)$$ making $$ \sqrt[n]{ \prod_{i=1}^{n}{(x_i+k)}} - k =\frac{\sum_{i=1}^{n} x_i}{n}+\frac{\left(\sum_{i=1}^{n} x_i\right)^2-n\sum_{i=1}^{n} x^2_i}{2n^2 k}+O\left(\frac{1}{k^2}\right)$$
For example, using $n=10$, $k=1000$ and $x_i=p_i$, the approximation would give $\frac{2572671}{200000}\approx 12.8634$ while the exact calculation would lead to $\approx 12.8639$
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You can also give an argument using L'Hopital's rule: \begin{align*} \lim_{k \to \infty} \sqrt[n]{\prod_{i = 1}^n (x_i + k)} - k &= \lim_{k \to \infty} \frac{\sqrt[n]{\prod_{i = 1}^n \left(\frac{x_i}{k} + 1\right)} - 1}{\frac{1}{k}} \\ &\overset{(H)}{=} \lim_{k \to \infty} \frac{\frac{1}{n} \cdot \frac{\sum_{i = 1}^n \frac{-x_i}{k^2} \cdot \prod_{j \ne i} \left(\frac{x_i}{k} + 1\right)}{\left(\prod_{i = 1}^n \left(\frac{x_i}{k} + 1\right)\right)^{1 - 1/n}}}{\frac{-1}{k^2}} \\ &= \lim_{k \to \infty}\frac{1}{n} \cdot \frac{\sum_{i = 1}^n x_i \cdot \prod_{j \ne i} \left(\frac{x_i}{k} + 1\right)}{\left(\prod_{i = 1}^n \left(\frac{x_i}{k} + 1\right)\right)^{1 - 1/n}} \\ &= \frac{\sum_{i = 1}^n x_i}{n} \end{align*}
On
$$ \begin{align} \lim_{k\to\infty}\left[\left(\prod_{i=1}^n(x_i+k)\right)^{1/n}-k\right] &=\lim_{k\to\infty}k\left[\prod_{i=1}^n\left(1+\frac{x_i}k\right)^{1/n}-1\right]\\ &=\lim_{k\to\infty}\frac{\prod\limits_{i=1}^n\left(1+\frac{x_i}k\right)^{1/n}-1}{\log\left(\prod\limits_{i=1}^n\left(1+\frac{x_i}k\right)^{1/n}\right)}\cdot\lim_{k\to\infty}\frac kn\sum_{i=1}^n\log\left(1+\frac{x_i}k\right)\\ &=\lim_{x\to1}\frac{x-1}{\log(x)}\cdot\frac1n\sum_{i=1}^n\lim_{k\to\infty}k\log\left(1+\frac{x_i}k\right)\\[9pt] &=\frac1n\sum_{i=1}^nx_i \end{align} $$
Assume that $k$ is greater than twice $|x_1|+\ldots+|x_n|$ and also greater than twice $M=x_1^2+\ldots+x_n^2$. Over the interval $\left(-\frac{1}{2},\frac{1}{2}\right)$ we have $$ e^x = 1+x+C(x)x^2, \qquad \log(1+x)=x+D(x)x^2 $$ with $|C(x)|,|D(x)|\leq 1$. It follows that
$$\begin{eqnarray*}\text{GM}(x_1+k,\ldots,x_n+k)&=&k\cdot \text{GM}\left(1+\tfrac{x_1}{k},\ldots,1+\tfrac{x_n}{k}\right)\\&=&k\exp\left[\frac{1}{n}\sum_{j=1}^{n}\log\left(1+\frac{x_j}{k}\right)\right]\\&=&k\exp\left[\frac{1}{k}\sum_{j=1}^{n}\frac{x_j}{n}+\Theta\left(\frac{M}{nk^2}\right)\right]\\&=&k\left[1+\frac{1}{k}\sum_{j=1}^{n}\frac{x_j}{n}+\Theta\left(\frac{M}{nk^2}\right)\right]\end{eqnarray*}$$ and the claim is proved. It looks very reasonable also without a formal proof: the magnitude of the difference between the arithmetic mean and the geometric mean is controlled by $\frac{\text{Var}(x_1,\ldots,x_n)}{\text{AM}(x_1,\ldots,x_n)}$. A translation towards the right leaves the variance unchanged and increases the mean.