# Weaker version of the $A.M.\;-\;G.M.$ inequality

It is a well known fact that, for $\mathbf{x} \in R^k$ with $\mathbf{x}=(x_1,...,x_k) \geq 0$, $A(\mathbf{x})=G(\mathbf{x})$ iff $x_1=x_2=...=x_n$, where $A(\mathbf{x})$ and $G(\mathbf{x})$ are the Arithmetic Mean and Geometric Mean of $x_1,...,x_k$ respectively. (Attainment of equality in AM-GM inequality).

My question is, does, for any $\epsilon>0$, $|A(\mathbf{x})-G(\mathbf{x})| <\epsilon \implies \sum_{i,j}|x_i-x_j|<c$ $\epsilon$ for some constant $c$, and vice versa.

To state in words: From the A.M-G.M. inequality condition, it is known that A.M. and G.M. are equal iff the individual terms are equal. Now, does this hold, that A.M. and G.M. are close enough, implies the individual terms are close enough to each other? and vice versa.

• – pharmine Mar 12 '15 at 9:42