An inequality about weighted AM-GM

If $a_i$ positive real numbers, $n \ge 2$ and $m$ positive integer, then I have to prove that $$\frac {(a_1^m a_2 + a_2^m a_3 + … + a_n^m a_1)^m}{(a_1^m + a_2^m + … + a_n^m)^{m+1}} \le \frac {1}{n} .$$ By Weighted AM-GM-HM inequalities, we have

$$\frac {a_1^m a_2 + a_2^m a_3 + … + a_n^m a_1}{a_1^m + a_2^m + … + a_n^m} \ge (a_2^{ a_1^m} a_3^{a_2^m} … a_1^{a_n^m})^{\frac {1}{ a_1^m + a_2^m + … + a_n^m }} \ge \frac { a_1^m + a_2^m + … + a_n^m }{\frac{ a_1^m }{a_2} + \frac{ a_2^m }{a_3}+ … + \frac{ a_n^m }{a_1} }$$

By AM-GM inequality we have

$$a_1^m + a_2^m + … + a_n^m \ge n \sqrt [n] { a_1^m a_2^m … a_n^m }$$

but I can’t go further. Thank you for your time

1 Answer

For $a_3=...=a_n\rightarrow0^+$ and $m=2$ we need to prove that $$(a_1^2+a_2^2)^3\geq na_1^4a_2^2,$$ which is wrong for the big $n$.

• are you sure? A friend claimed that is right, I don't know Apr 19, 2018 at 17:29
• I am sure. Your inequality is wrong. See my post and $a_1=a_2=1$, $n=9$. Apr 19, 2018 at 17:31
• but for $a_1=...=a_n=1$ for example, is true Apr 19, 2018 at 17:34