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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:

Plot[f[x, 2, 2], {x, 1/2, 4}]

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?

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up vote 7 down vote accepted

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 $2$.

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Thanks! This means that $b^2$ needs to be replaced by $2b^2$ in the left hand side. Same for $a^2$ in the right hand side, right? – ivan Sep 21 '12 at 9:43
@ivan: Right. And you're welcome. – joriki Sep 21 '12 at 9:46
I've dropped an email to Prof. Dragomir about this. – ivan Sep 21 '12 at 10:59

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