Show inequality generalization $\sum (x_i-1)(x_i-3)/(x_i^2+3)\ge 0$ 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?

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
$$
\sum_{k=0}^7f(x_k)=-0.111270522834348
$$

Explanation:
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.
$\hspace{2cm}$
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 $0$.
