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

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?

• Just an observation, with $a_1=2~\&~a_2=\frac 1 2$, this would be false for $1<\alpha<2.3$. Intuitively the summand minus 1 as a function of $\ln a_i$ is not strictly curving upwards for all $\alpha$ up to $3$ so the inequality should be false, but I don't know
– user632577
May 29, 2019 at 11:04

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 $$\alpha\in(2-\alpha_0,1-\sqrt{2}]\cup[1+\sqrt{2},\alpha_0)$$.