Given $0 < y < 1 < n \in \mathbb Z^+$ and $n$ positives $x_1, x_2, \cdots, x_{n – 1}, x_n$ such that $\displaystyle \sum_{i = 1}^nx_i = n$.
Prove that $$\large \frac{n}{3} \le \sum_{i = 1}^n\frac{(n – y)^2}{y^2(x_i^2 + x_i + 1) – 3ny(x_i + 1) + 3n^2} \le \sum_{i = 1}^n\frac{1}{x_i^2 + x_i + 1}$$
We have that
$$\sum_{i = 1}^n\frac{(n – y)^2}{y^2(x_i^2 + x_i + 1) – 3ny(x_i + 1) + 3n^2} – \frac{n}{3}$$
$$\frac{y}{3} \cdot \sum_{i = 1}^n\frac{(x_i – 1)(2n – x_iy – 2y)}{x_i^2y^2 + x_iy^2 + y^2 – 3nx_iy – 3ny + 3n^2}$$
and
$$\sum_{i = 1}^n\frac{1}{x_i^2 + x_i + 1} – \sum_{i = 1}^n\frac{(n – y)^2}{y^2(x_i^2 + x_i + 1) – 3ny(x_i + 1) + 3n^2}$$
$$ = n \cdot \sum_{i = 1}^n\frac{(x_i – 1)(nx_i + 2n – 2x_iy – y)}{(x_i^2 + x_i + 1)(x_i^2y^2 + x_iy^2 + y^2 – 3nx_iy – 3ny + 3n^2)}$$
So that's no help.
Besides, $$\sum_{i = 1}^n\frac{(n – y)^2}{y^2(x_i^2 + x_i + 1) – 3ny(x_i + 1) + 3n^2}$$ is a concave function.
So Jensen's inequality could help here, but I don't know how.
Best Answer
The left inequality.
We need to prove that $$\sum_{i=1}^n\left(\frac{(n - y)^2}{y^2(x_i^2 + x_i + 1) - 3ny(x_i + 1) + 3n^2}-\frac{1}{3}\right)\geq0$$ or $$\sum_{i=1}^n\frac{(x_i-1)(3n-2y-yx_i)}{y^2(x_i^2 + x_i + 1) - 3ny(x_i + 1) + 3n^2}\geq0$$ or $$\sum_{i=1}^n\left(\frac{(x_i-1)(3n-2y-yx_i)}{y^2(x_i^2 + x_i + 1) - 3ny(x_i + 1) + 3n^2}-\frac{x_i-1}{n-y}\right)\geq0$$ or $$\sum_{i=1}^n\frac{(x_i-1)^2(2n-y-yx_i)}{y^2(x_i^2 + x_i + 1) - 3ny(x_i + 1) + 3n^2}\geq0,$$ which is obvious.
For $n\geq3$ we can prove a right inequality by the same way.
Also, easy to check that the right inequality is true for $n=2$.