# General inequality

Hello I would like to prove this

My strategy (with $n\geq 3$):

It's simple we have this inequality from this link (p.7)

If $a_1,a_2,...,a_n$ are nonnegative real numbers such that $a_1+\cdots+a_n=n$ then :

$$\frac{a_1^{n+1}+\cdots+a_n^{n+1}-n}{a_1^{2}+\cdots+a_n^{2}-n}\geq (n-1)\left[\left(\frac{n}{n-1}\right)^n-1\right]$$

Or:

$$\frac{a_1^{n+1}+\cdots+a_n^{n+1}-n}{(n-1)(a_1^{2}+\cdots+a_n^{2}-n)}+1\geq \left(\frac{n}{n-1}\right)^n$$

On the other hand we have with $a_1+\cdots+a_n=n$ and $a_1,...,a_n$ positive real numbers :

$$\left(\frac{a_1}{a_2}+\frac{a_2}{a_3}+\cdots+\frac{a_n}{a_1}-n\right)\leq\left(\frac{n}{n-1}\right)^n\left(\frac{1}{a_1a_2\cdots a_n }-1\right)$$

Or :

$$\left(\frac{1}{a_1a_2\cdots a_n }-1\right)\left(\frac{a_1}{a_2}+\frac{a_2}{a_3}+\cdots+\frac{a_n}{a_1}-n\right)^{-1}\geq \left(\frac{n-1}{n}\right)^n$$

So if we multiply this two inequality we have to prove this to get the desired result :

$$\left[\frac{a_1^{n+1}+\cdots+a_n^{n+1}-n}{(n-1)(a_1^{2}+\cdots+a_n^{2}-n)}+1\right]\left(\frac{1}{a_1a_2\cdots a_n }-1\right)\left(\frac{a_1}{a_2}+\frac{a_2}{a_3}+\cdots+\frac{a_n}{a_1}-n\right)^{-1}\geq 1.$$

And conclude by deduction .

My question is : how to prove this last inequality with the conditions mentioned above ?

My partial answer : I think we can use a generalized Radon-type inequality .

• For the first inequality we can use EV method. For the second inequality we can't do it. I think. – Michael Rozenberg Aug 7 '17 at 16:00

we have with $a_1+\cdots+a_n=n$ and $a_1,...,a_n$ positive real numbers :
$$\left(\frac{a_1}{a_2}+\frac{a_2}{a_3}+\cdots+\frac{a_n}{a_1}-n\right)\leq\left(\frac{n}{n-1}\right)^n\left(\frac{1}{a_1a_2\cdots a_n }-1\right)$$
It is easy to check that this inequality holds for $n=2$. But it fails already for $n=3$. Indeed, let $a_1=\frac 32$, $a_2=\frac 12$, and $a_3=1$. Then its left hand side equals $\frac 76$, whereas its right hand side equals $\frac 98$.