Inequality concerning factorial 

*

*$a_1,a_2\cdots,a_n\in\mathbb R_+$;

*$\forall1\le k\le n,a_1a_2\cdots a_k\ge k!$
Show that $$\frac{2!}{1+a_1}+\frac{3!}{(1+a_1)(2+a_2)}+\cdots+\frac{(n+1)!}{(1+a_1)(2+a_2)\cdots(n+a_n)}<3.$$

My thought would be to prove $$(1+a_1)(2+a_2)\cdots(n+a_n)\ge2^nn!.$$ If so, the expression on the left $<\sum_{i=1}^n\frac{(i+1)!}{2^ii!}=\sum_{i=1}^n\frac{i+1}{2^i}=3-\frac{3+n}{2^n}<3$ and that’s done.
Perhaps we could find an upper limit for it and use the “spiral induction” method?
 A: Your hypothesis is correct! We can indeed show that
$$\prod_{i=1}^n\left(\frac{1}{2}i+\frac{1}{2}a_i\right)\geq n!$$
To do this, we can show the following properties of $f(x)=x_1x_2\cdots x_n$:

*

*$f$ is quasi-concave on $\mathbb{R}_+^n$. This follows from log-concavity of $f$ which itself is true since
$$\ln(f(x)) = \ln(x_1) + \cdots + \ln(x_2)$$
and $\ln(x)$ is concave.

*$f(1,2,\cdots,n) = n!$

*$f(a_1,a_2,\cdots,a_n)\geq n!$
Then, since $f$ is quasiconcave, the upper-level set $\{x\in \mathbb{R}_+^n : f(x) \geq z \}$ is convex on $\mathbb{R}_+^n$.
Finally, taking a convex combination of $(1,2,\cdots,n)$ and $(a_1,a_2,\cdots,a_n)$, the proof is complete.
A: Your thoughts are indeed correct, thank you to @IsaacBrowne for a solution involving concavity. It is also possible to use more elementary methods with the AM-GM inequality as $x+y\geq2\sqrt{xy}$ for positive $x,y$:
$$\prod_{i=1}^{n}(a_i+i)\geq \prod_{i=1}^{n}2\sqrt{ia_i}=\sqrt{i!}2^n\prod_{i=1}^n\sqrt{a_i}\geq \sqrt{i!}2^n \sqrt{i!}=i!2^n$$
As a side note, what is spiral induction? I keep on hearing people mention it but no idea what it is, and searches never reveal anything.
