Is this AM/GM refinement correct or not? In Chap 1.22 of their book Mathematical Inequalities, Cerone and Dragomir prove the following interesting inequality. Let $A_n(p,x)$ and $G_n(p,x)$ denote resp. the weighted arithmetic and the weighted geometric means, where $x_i\in[a,b]$ and $p_i\ge0$. $P_n$ is the sum of all $p_i$. Then the following holds:
$$
\exp\left[\frac{1}{b^2P_n^2}\sum\limits_{i<j} p_ip_j(x_i-x_j)^2\right]\le\frac{A_n(p,x)}{G_n(p,x)} \le\exp\left[\frac{1}{a^2P_n^2}\sum\limits_{i<j} p_ip_j(x_i-x_j)^2\right]
$$
The relevant two pages of the book may be consulted here.
I need help to figure out what is wrong with my next arguments. I will only be interested in the LHS of the inequality. Let $n=3$ and let $p_i=1$ for all $i$ and hence $P_n=3$. Let $x,y,z\in[a,b]$. We can assume that $b=\max\{x,y,z\}$. Our inequality is equivalent to:
$$
f(x,y,z)=\frac{x+y+z}{3\sqrt[3]{xyz}}-\exp\left[\frac{(x-y)^2+(x-z)^2+(y-z)^2}{9\max\{x,y,z\}^2}\right]\ge0
$$
According to Mathematica $f(1, 2, 2)=-0.007193536514508<0$ which means that the inequality as stated is incorrect. Moreover, if I plot the values of $f(x,2,2)$ here is what I get:

You can download my Mathematica notebook here.
As you can see our function is negative for some values of $x$ which means that the inequality does not hold for those values.
Obviously it is either me that is wrong or Cerone and Dragomir's derivation. I have read their proofs and I can't find anything wrong so I suspect there is a flaw in my exposition above. 
Can someone help me find it?

Your modesty in suspecting that the error is yours is commendable, but in fact you found an error in the book. The "simple calculation" on p. $49$ is off by a factor of $2$, as you can easily check using $n=2$ and $p_1=p_2=1$. Including a factor $\frac12$ in the inequality makes it come out right.
You can also check this by using $f(x)=x^2$, $n=2$, $p_1=p_2=1$ and $x_1=-1$, $x_2=1$ in inequality $(1.151)$ on p. $48$. Then the difference between the average of the function values and the function value of the average is $1$, and the book's version of the inequality says that it's $2$.
