\begin{gather\*}
M\_g=(x\_1\times x\_2\times\dotsb\times x\_n)^{1/n} \\
M\_a=\frac1 n\times (x\_1+x\_2+\dotsb+x\_n).
\end{gather\*}


Is it true that $$\lvert M\_g-M\_a\rvert \leq (\max(x\_i) /\min(x\_i)) \times(\max(x\_i) - \min(x\_i))?\label{1}\tag{1}$$


And is it true that


$$\lvert M\_g-M\_a\rvert\leq (\max(x\_i) /\min(x\_i)) \times\left( \frac1{n^2}\sum \limits \_{(i, j) \in \{1,\dotsc,n\}^2} \lvert x\_i-x\_j\rvert\right)?\label{2}\tag{2}$$


Ps.: $x\_i\in\mathbb R\_+^\*$.



Compare AM and GM
The inequality $(2)$ (even with factor $\frac12$ in the r.h.s.) follows from the inequality quoted in [this answer](https://mathoverflow.net/q/215094):
$$M\_a - M\_g \leq \frac1{2n\min\_k x\_k} \sum\_{i=1}^n (x\_i - M\_a)^2.$$
First, we notice that
\begin{split}
(x\_i - M\_a)^2 &\leq \max\_k x\_k\cdot |x\_i-M\_a|\\
&= \max\_k x\_k\cdot \left|\frac1n \sum\_{j=1}^n (x\_i - x\_j)\right| \\
& \leq \frac{\max\_k x\_k}n \sum\_{j=1}^n |x\_i - x\_j|.
\end{split}
Then
$$\frac1{2n\min\_k x\_k} \sum\_{i=1}^n (x\_i - M\_a)^2 \leq \frac12 \frac{\max\_k x\_k}{\min\_k x\_k} \frac1{n^2} \sum\_{i,j=1}^n |x\_i - x\_j|.$$
QED


