\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.: $x_i\in\mathbb R_+^*$.
Best Answer
The inequality $(2)$ (even with factor $\frac12$ in the r.h.s.) follows from the inequality quoted in this answer: $$M_a - M_g \leq \frac1{2n\min_k x_k} \sum_{i=1}^n (x_i - M_a)^2.$$ 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