$\prod_{i=1}^{n-1} a_i = 1 \Rightarrow \prod_{i=1}^{n-1} (1+ a_i)^{i+1} > n^n$?

Question

Let $n>3$ be an integer number and $a_1, a_2, \dots, a_{n-1}$ positive real numbers, such that $\prod_{i=1}^{n-1} a_i = 1$.

Is the following inequality true? $$ \prod_{i=1}^{n-1} (1+ a_i)^{i+1} > n^n $$

Answer 1

You can prove it as follows using AM-GM: $$ \prod_{i=1}^{n-1}(1+a_i)^{i+1}=\prod_{i=1}^{n-1}\left((i+1)\left(\frac{i\cdot\frac{1}{i}+a_i}{i+1}\right)\right)^{i+1}\ge\prod_{i=1}^{n-1}\left((i+1)\left(\frac{a_i}{i^{i}}\right)^{\frac{1}{i+1}}\right)^{i+1}=\prod_{i=1}^{n-1}\left(\frac{(i+1)^{i+1}}{i^{i}}\cdot a_i\right)=\frac{2^2\cdot3^3\cdots (n-1)^{n-1}\cdot n^n}{1^1\cdot2^2\cdots (n-1)^{n-1}}\cdot\prod_{i=1}^{n-1}a_i=n^n $$ Furthermore, the inequality is sharp, because for equality, we would need: $$ a_i=\frac{1}{i}\space\forall i\in\{1,2,\dots,n-1\} $$ Which is a contradiction if $n\ge3$ because then: $$ \prod_{i=1}^{n-1}a_i=\frac{1}{(n-1)!}<1 $$ And therefore: $$ \prod_{i=1}^{n-1}(1+a_i)^{i+1}> n^n $$