Prove that $\frac{n}{3} \le \sum_{i = 1}^n\frac{(n - y)^2}{y^2(x_i^2 + x_i + 1) - 3ny(x_i + 1) + 3n^2} \le \sum_{i = 1}^n\frac{1}{x_i^2 + x_i + 1}$.

Question

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.

Answer 1

Proof of the right inequality.

For all $n\geq3$ we obtain: $$\sum_{i=1}^n\frac{1}{x_i^2+x_i+1}-\sum_{i=1}^n\frac{(n - y)^2}{y^2(x_i^2 + x_i + 1) - 3ny(x_i + 1) + 3n^2}=$$ $$=\sum_{i=1}^n\left(\frac{1}{x_i^2+x_i+1}-\frac{(n - y)^2}{y^2(x_i^2 + x_i + 1) - 3ny(x_i + 1) + 3n^2}\right)=$$

$$=n\sum_{i=1}^n\frac{(1-x_i)((n-2y)x_i+2n-y)}{(x_i^2+x_i+1)(y^2(x_i^2 + x_i + 1) - 3ny(x_i + 1) + 3n^2)}=$$ $$=n\sum_{i=1}^n\left(\frac{(1-x_i)((n-2y)x_i+2n-y)}{(x_i^2+x_i+1)(y^2(x_i^2 + x_i + 1) - 3ny(x_i + 1) + 3n^2)}+\frac{x_i-1}{3(n-y)}\right)=$$ $$=\frac{n}{3(n-y)}\sum_{i=1}^n\frac{(x_i-1)^2(y^2x_i^3-3y(n-y)x_i^2+3(n-y)(n-2y)x_i+3n^2-6ny+2y^2)}{(x_i^2+x_i+1)(y^2(x_i^2 + x_i + 1) - 3ny(x_i + 1) + 3n^2)}.$$ We'll prove that $$y^2x^3-3y(n-y)x^2+3(n-y)(n-2y)x+3n^2-6ny+2y^2\geq0$$ is true for any $n\geq3$, $x>0$ and $0<y<1.$

Let $\dfrac{n}{y}=t$.

Thus, $t>3$ and we need to prove that $f(x)\geq0,$ where $$f(x)=x^3-3(t-1)x^2+3(t^2-3t+2)x+3t^2-6t+2.$$ Indeed, $$f'(x)=3x^2-6(t-1)x+3(t^2-3t+2),$$ which gives $$x_{min}=t-1+\sqrt{(t-1)^2-(t^2-3t+2)}=t-1+\sqrt{t-1},$$ which gives $$f(x)\geq f\left(t-1+\sqrt{t-1}\right)=\left(\sqrt{(t-1)^3}-1\right)^2-2>0.$$ Id est, it's enough to prove our inequality for $n=2$.

Let $x_1=a$.

Thus, $x_2=2-a$, $0<a<2$ and we need to prove that $$\tfrac{1}{a^2+a+1}+\tfrac{1}{(2-a)^2+2-a+1}\geq\tfrac{(2 - y)^2}{y^2(a^2 + a + 1) - 6y(a + 1) + 12}+\tfrac{(2 - y)^2}{y^2((2-a)^2 + 2-a + 1) - 6y(2-a + 1) + 12}$$ or $$(1-y)(a-1)^2(y^2(11+16a-12a^2+4a^3-a^4)+12(1-y)(5+2a-a^2))\geq0,$$ for which it's enough to prove that $$11+16a-12a^2+4a^3-a^4\geq0,$$ which is true by AM-GM: $$11+16a-12a^2+4a^3-a^4\geq11+16a-12a^2+2a^3=$$ $$=11+2\cdot8a+4\cdot\frac{a^3}{2}-12a^2\geq\left(7\sqrt[7]{11\cdot8^2\cdot\left(\frac{1}{2}\right)^4}-12\right)a^2>0.$$

Answer 2

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 $n=2$.