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?

• @Wizard0001, I think that violates the condition that $a_1a_2…a_k\geq k!$ as $0\times 0<2!$ May 17 at 17:53

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.

• Spiral induction is like this. We want $f(x)<g(x),x\in\mathbb{N}_+$ and to prove it we strengthen the proposition to$f(x)<g(x)<h(x)$. If $f(x)<g(x)\Rightarrow g(x+1)<h(x+1);g(x)<h(x)\Rightarrow f(x+1)<g(x+1)$ we can prove the original problem. May 18 at 3:39
• Found it here: baike.baidu.com/item/螺旋式归纳法/11064725 but it’s Chinese so perhaps you could read it with a browser that can translate web pages. May 18 at 4:48
• Actually that works, luckily I can read Chinese, thanks! May 18 at 4:58

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 $$f(x)=x_1x_2\cdots x_n$$:

1. $$f$$ 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.
2. $$f(1,2,\cdots,n) = n!$$
3. $$f(a_1,a_2,\cdots,a_n)\geq n!$$

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.
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