The powers are n-1 and the last index is n-1

VicKmath7 wrote: P4 For positive real numbers $a_1, a_2, ..., a_n$ with product 1 prove: $(a_1/a_2)^(n-1)+(a_2/a_3)^(n-1)+...+(a_(n-1)/a_n)^(n-1)=>a_1^2+a_2^2+...+a_n^2$ For positive real numbers $a_1, a_2, ..., a_n$ such that $\prod_{i=1}^{n} a_i = 1$. Prove: $$\left(\frac{a_1}{a_2}\right)^{n-1}+\left(\frac{a_2}{a_3}\right)^{n-1}+...+\left(\frac{a_{n-1}}{a_n}\right)^{n-1} \geq a_1^{2}+a_2^{2}+...+a_n^{2}$$

bel.jad5 wrote: VicKmath7 wrote: P4 For positive real numbers $a_1, a_2, ..., a_n$ with product 1 prove: $(a_1/a_2)^(n-1)+(a_2/a_3)^(n-1)+...+(a_(n-1)/a_n)^(n-1)=>a_1^2+a_2^2+...+a_n^2$ Do you mean? $$\left(\frac{a_1}{a_2}\right)^{n-1}+\left(\frac{a_2}{a_3}\right)^{n-1}+...+\left(\frac{a_{n-1}}{a_n}\right)^{n-1} \geq a_1^{2}+a_2^{2}+...+a_n^{2}$$ $ a_1=a_2=...=a_n $ $n>na^2 ??? $

Feridimo wrote: bel.jad5 wrote: VicKmath7 wrote: P4 For positive real numbers $a_1, a_2, ..., a_n$ with product 1 prove: $(a_1/a_2)^(n-1)+(a_2/a_3)^(n-1)+...+(a_(n-1)/a_n)^(n-1)=>a_1^2+a_2^2+...+a_n^2$ Do you mean? $$\left(\frac{a_1}{a_2}\right)^{n-1}+\left(\frac{a_2}{a_3}\right)^{n-1}+...+\left(\frac{a_{n-1}}{a_n}\right)^{n-1} \geq a_1^{2}+a_2^{2}+...+a_n^{2}$$ $ a_1=a_2=...=a_n $ $n>na^2 ??? $ He said $\prod_{i=1}^{n}a_i = 1$

bel.jad5 wrote: VicKmath7 wrote: P4 For positive real numbers $a_1, a_2, ..., a_n$ with product 1 prove: $(a_1/a_2)^(n-1)+(a_2/a_3)^(n-1)+...+(a_(n-1)/a_n)^(n-1)=>a_1^2+a_2^2+...+a_n^2$ For positive real numbers $a_1, a_2, ..., a_n$ such that $\prod_{i=1}^{n} a_i = 1$. Prove: $$\left(\frac{a_1}{a_2}\right)^{n-1}+\left(\frac{a_2}{a_3}\right)^{n-1}+...+\left(\frac{a_{n-1}}{a_n}\right)^{n-1} \geq a_1^{2}+a_2^{2}+...+a_n^{2}$$ Thanks Everything is clear to me now

bel.jad5 wrote: Feridimo wrote: bel.jad5 wrote: Do you mean? $$\left(\frac{a_1}{a_2}\right)^{n-1}+\left(\frac{a_2}{a_3}\right)^{n-1}+...+\left(\frac{a_{n-1}}{a_n}\right)^{n-1} \geq a_1^{2}+a_2^{2}+...+a_n^{2}$$ $ a_1=a_2=...=a_n $ $n>na^2 ??? $ He said $\prod_{i=1}^{n}a_i = 1$ Still doesn't make sense since if all the $a_i = 1$, then we have $n-1\ge n$. Maybe the left hand side is cyclic with the last term being $\left(\frac{a_n}{a_1}\right)^{n-1}$?

Thank you Ivannavl, corrected

Using weighted A.M.-G.M., it's enough to find an $n$-tuple of non-negative real numbers $e_1,e_2,\dotsc ,e_n$ such that $e_1+e_2+\cdots +e_n=1$, $e_1-e_n=(2-2/n)/(n-1)=2/n$, and $e_k-e_{k-1}=(-2/n)/(n-1)$ for all $k=2,3,\dotsc ,n$. For this end, simply choose $e_k=2(n-k)/(n(n-1))$ for all $k=1,2,\dotsc ,n$.

Let $a_1, a_2, \cdots , a_n$ be positive real numbers such that $ a_1 a_2 \cdots a_n= 1.$ Prove that $$\left(\frac{a_1}{a_2}\right)^{n-1}+\left(\frac{a_2}{a_3}\right)^{n-1}+\cdots+\left(\frac{a_{n-1}}{a_n}\right)^{n-1}+\left(\frac{a_n}{a_1}\right)^{n-1} \geq a_1^2+a_2^2+\cdots+a_n^2+\frac{1}{a_1^2}+\frac{1}{a_2^2}+\cdots+\frac{1}{a_n^2}-n.$$

sqing wrote: Let $a_1, a_2, \cdots , a_n$ be positive real numbers such that $ a_1 a_2 \cdots a_n= 1.$ Prove that $$\left(\frac{a_1}{a_2}\right)^{n-1}+\left(\frac{a_2}{a_3}\right)^{n-1}+\cdots+\left(\frac{a_{n-1}}{a_n}\right)^{n-1}+\left(\frac{a_n}{a_1}\right)^{n-1} \geq a_1^2+a_2^2+\cdots+a_n^2+\frac{1}{a_1^2}+\frac{1}{a_2^2}+\cdots+\frac{1}{a_n^2}-n.$$ Do you have a proof or are you proposing this inequality?

The official solution can be found here: https://pregatirematematicaolimpiadejuniori.files.wordpress.com/2020/03/stars_j_2019.pdf