inequality with $a_{1}a_{2}\cdots a_{n}=1$

Question

let $a_{i}>0$,and such $$a_{1}a_{2}\cdots a_{n}=1$$ find all real numbers $\alpha$,then $$\sum_{i=1}^{n}\dfrac{a^{\alpha}_{i}}{a^2_{i}-a_{i}+1}\ge n$$

I have solve $\alpha\ge 3$,But I can't sure $\alpha>1$ this inequality right?

Answer 1

Just some incomplete thoughts.

By playing around with some plots, we can see that $$ \frac{x^\alpha}{x^2-x+1}\geq 1+(\alpha-1)\log(x) $$ for all $x>0$ and $\alpha\geq\alpha_0$ where $\alpha_0\approx 2.7461845658404$. Clearly this then immediately implies your original inequality. Also we see that if the inequality works for some $\alpha$, then it also works for $2-\alpha$ (just replace each $a_i$ by $\frac{1}{a_i}$).

On the other hand, with $(a_1,…,a_n)=(x,\frac{1}{x},1,…,1)$, the left hand side of your inequality is equal to $$ \frac{x^\alpha}{x^2-x+1}+\frac{x^{-\alpha}}{x^{-2}-x^{-1}+1}+n-2=\frac{x^\alpha+x^{2-\alpha}}{x^2-x+1}+n-2 $$ so the inequality for this tupel is equivalent to $$ x^\alpha+x^{2-\alpha}\geq 2(x^2-x+1). $$ As we have equality for $x=1$ and also the derivatives of the two sides agree at $x=1$, we must have that the second derivative of the left hand side at $x=1$ is greater than or equal to the second derivative of the right hand side at $x=1$. By calculation, this then amounts to $2(\alpha-1)^2\geq 4$, i.e. either $\alpha\geq 1+\sqrt{2}\approx 2.414$, or $\alpha\leq 1-\sqrt{2}\approx -0.586$.

So in summary the inequality holds for all $\alpha\in (-\infty,2-\alpha_0]\cup[\alpha_0,\infty)$ (with $\alpha_0\approx 2.7461845658404$), and it doesn’t hold in general for all $\alpha\in (1-\sqrt{2},1+\sqrt{2})$. So it remains to see what happens for $\alpha\in(2-\alpha_0,1-\sqrt{2}]\cup[1+\sqrt{2},\alpha_0)$.