Knowing that $\prod_{i = 1}^na_i = 1$, prove that $\prod_{i = 1}^n(a_i + 1)^{i + 1} > (n + 1)^{n + 1}$. Given natural $n$ $(n \ge 3)$ and positives $a_1, a_2, \cdots, a_{n - 1}, a_n$ such that $\displaystyle \prod_{i = 1}^na_i = 1$, prove that $$\large \prod_{i = 1}^n(a_i + 1)^{i + 1} &gt; (n + 1)^{n + 1}$$


We have that $$\prod_{i = 1}^n(a_i + 1)^{i + 1} \ge \prod_{i = 1}^n(2\sqrt{a_i}) \cdot \left(\sqrt[m]{\prod_{i = 1}^na_i^i} + 1\right)^m$$

where $\displaystyle p = \sum_{i = 1}^ni = \dfrac{n(n + 1)}{2}$, then I don't know what to do next.

I suspect that $\displaystyle \min\left(\prod_{i = 1}^n(a_i + 1)^{i + 1}\right) = 2^q$, occuring when $a_1 = a_2 = \cdots = a_{n - 1} = a_n = 1$, where $q = \dfrac{(n + 3)n}{2}$, although I'm not sure that $2^q &gt; (n + 1)^{n + 1}, \forall n \in \mathbb Z^+, n \ge 2$.

(I've just realised this is just a redraft of problem 2, IMO 2012.)
For the numbers $2^{\frac{n}{2}}$ is greater than n+1 obviously satisfies, for n>5 can be proven by induction (hint: x$\sqrt{2}$ - x - 1>0 for x> $\sqrt{2}$ + 1)and for other n&lt;6 check manually
According to the AM-GM inequality and telescopic products, we have that $$\prod_{i = 1}^n(a_i + 1)^{i + 1} = \prod_{i = 1}^n\left(a_i + i \cdot \frac{1}{i}\right)^{i + 1} \ge \prod_{i = 1}^n\left[a_i \cdot \frac{(i + 1)^{i + 1}}{i^i}\right] = (i + 1)^{i + 1}$$

The equality sign occurs when $\displaystyle a_i = \dfrac{1}{i},i = \overline{1, n} \implies \prod_{i = 1}^na_i = \prod_{i = 1}^n\dfrac{1}{i} \implies \prod_{i = 1}^ni = 1$, which is false, $\forall n \in \mathbb N, n \ge 2$.

$$\implies \prod_{i = 1}^n(a_i + 1)^{i + 1} &gt; (i + 1)^{i + 1}$$