We have to prove \[(n+1)(a_{1}+1)(2a_{2}+1)\cdot..\cdot(na_{n}+1)\geq 2^{n}(1+a_{1}+2a_{2}+...+na_{n}).\] Denoting $x_{i}=ia_{i}\geq 1$, we get \[(n+1)(x_{1}+1)...(x_{n}+1)\geq 2^{n}(1+x_{1}+...+x_{n})\] Now use mathematical induction : $n=1$, $2(x_{1}+1)\geq 2(1+x_{1})$ Assume \[n(x_{1}+1)...(x_{n}+1)\geq 2^{n-1}(1+x_{1}+...+x_{n-1})\] Denote $S=1+x_{1}+...+x_{n-1}\geq n$ It is enough to prove $(n+1)S(x_{n}+1)\geq 2n(S+x_{n}),$ which is reduces to $nS(x_{n}-1)+2x_{n}(S-n)-S(x_{n}-1)\geq 0$ or $(x_{n}-1)S(n-1)+2x_{n}(S-n)\geq 0.$ $\text{q.e.d.}$

HI, prowler! I have made the substitutions. However, I didn't use induction! I have considered the expression as a linear function in $x_{1}$ with positive "leading coefficient" (that is, an increasing linear function on $[1,\infty)$. So $x_{1}=1$. Repeating this step for each $i$ we get $x_{1}=x_{2}=\dots=x_{n}=1$. Good luck!

Let just induct Obviously, the inequality should hold for $n=1$. Assume it holds for $n=k$. We will prove it holds for $n=k+1$. Consider \[ LHS=\left(a_{1}+1\right)\left(a_{2}+\frac{1}{2}\right)\cdots\left(a_{n}+\frac{1}{n}\right)\left(a_{n+1}+\frac{1}{n+1}\right)\geq\frac{2^{n}}{(n+1)!}(1+a_{1}+2a_{2}+\dots+na_{n})\left(a_{n+1}+\frac{1}{n+1}\right) \]We will prove that \[ \begin{aligned} (1+a_{1}+2a_{2}+\dots+na_{n})\left(a_{n+1}+\frac{1}{n+1}\right) & \geq\frac{2}{n+2}(1+a_1+2a_2+\cdots+na_n)+\frac{2(n+1)a_{n+1}}{n+2}\\ \Leftrightarrow (1+a_{1}+2a_{2}+\dots+na_{n})\left(\frac{(a_{n+1}(n+1)(n+2)+n+2-2n-2}{(n+1)(n+2)}\right) & \geq\frac{2(n+1)a_{n+1}}{n+2} \end{aligned} \]By the given condition, we have $LHS\geq\frac{a_{n+1}(n+1)(n+2)-n}{n+2}$. From here, by simple algebraic manipulation, we got $a_{n+1}\geq\frac{1}{n+1}$, which is true.