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Let $f(x)=\dfrac{(x-1)(x-3)}{x^2+3}$. It seems to be that:

If $x_1,x_2,\ldots,x_n$ are positive real numbers with $\prod_{i=1}^n x_i=1$ then $\sum_{i=1}^n f(x_i)\ge 0$.

For $n>2$ a simple algebraic approach gets messy. This would lead to a generalization of this inequality, but even the calculus solution offered there for $n=3$ went into cases.

I thought about Jensen's inequality, but $f$ is not convex on $x>0$.

Can someone prove or disprove the above claim?

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Perhaps you want the condition to be $\sum_{i=1}^nf(x_i)\geq0$? Because letting every $x_i=1$ makes $f(x_i)=0$, and hence so also their sum. – Zev Chonoles Jul 8 '12 at 23:42
@Zev, yes, precisely, thanks. – Zander Jul 8 '12 at 23:43
up vote 1 down vote accepted

Unfortunately, this is not true.

Simple counterexample:

My original counterexample had some ugly numbers in it, but fortunately, there is a counterexample with nicer numbers. However, the explanation below might still prove informative.

Note that for $x>0$, $$ f(x)=\frac{(x-1)(x-3)}{x^2+3}=1-\frac{4x}{x^2+3}\lt1\tag{1} $$ Next, we compute $$ f(2)=-\frac17\tag{2} $$ Let $x_0=\frac{1}{256}$ and $x_k=2$ for $1\le k\le8$.

The product of the $x_k$ is $\frac{1}{256}\cdot2^8=1$, yet by $(1)$ and $(2)$, the sum of the $f(x_k)$ is less than $1-\frac87\lt0$.

Original couunterexample:

Let $x_0=e^{-3.85}$ and $x_k=e^{.55}$ for $1\le k\le 7$.

We get $f(x_0)=0.971631300121646$ and $f(x_k)=-0.154700260422285$ for $1\le k\le 7$.

Then, $$ \prod_{k=0}^7x_k=1 $$ yet $$ \sum_{k=0}^7f(x_k)=-0.111270522834348 $$


Let me explain how I came up with this example.

$\prod\limits_{k=0}^nx_k=1$ is equivalent to $\sum\limits_{k=0}^n\log(x_k)=0$. Therefore I considered $u_k=\log(x_k)$. Now we want $$ \sum_{k=0}^nu_k=0 $$ to mean that $$ \sum_{k=0}^n\frac{(e^{u_i}-1)(e^{u_i}-3)}{e^{2u_i}+3}\ge0 $$ I first looked at the graph of $\large\frac{(e^{u}-1)(e^{u}-3)}{e^{2u}+3}$. If the graph were convex, the result would be true.

$\hspace{2cm}$enter image description here

Unfortunately, the graph was not convex, but I did note that $f(u)$ dipped below $0$ with a minimum of less than $-\frac17$ near $u=.55$, and that it was less than $1$ everywhere. Thus, if I took $u=.55$ for $7$ points and $u=-3.85$ for the other, the sum of the $u_k$ would be $0$, yet the sum of the $f(e^{u_k})$ would be less than $0$.

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