Prove that $\tfrac{a_1a_2\cdots a_n(1-a_1-a_2-\cdots-a_n)}{(a_1+a_2+\cdots+a_n)(1-a_1)\cdots(1-a_n)} \leq \frac{1}{n^{n+1}}.$

Question

Let $a_1,a_2,\ldots,a_n$ be positive such that $a_1+a_2+\cdots+a_n < 1$. Prove that $$\dfrac{a_1a_2\cdots a_n(1-a_1-a_2-\cdots-a_n)}{(a_1+a_2+\cdots+a_n)(1-a_1)\cdots(1-a_n)} \leq \dfrac{1}{n^{n+1}}.$$

Attempt

I think I would have to prove this by induction, but I am unsure how to prove the induction. Proving the base case would just be $\dfrac{a_1(1-a_1)}{a_1(1-a_1)} = 1 \leq \dfrac{1}{1}$. For the inductive hypothesis I would assume $\dfrac{a_1a_2\cdots a_k(1-a_1-a_2-\cdots-a_k)}{(a_1+a_2+\cdots+a_k)(1-a_1)\cdots(1-a_k)} \leq \dfrac{1}{k^{k+1}}$ holds for some $k$. Then I would have to show it holds true for the $k+1$ case. How would I do that?

Is this somehow related to http://math.stackexchange.com/questions/1592849/prove-that-x-1x-2-cdots-x-n-geq-n-1n? — Martin R, Dec 29 '15 at 21:11

score 1 · Answer 1 · answered Dec 29 '15 at 21:23

Let $a_0:=1-\sum_{i=1}^n\,a_i$. Then, the inequality is equivalent to $$\prod_{i=0}^n\,\frac{a_i}{\sum_{j\in\{0,1,2,\ldots,n\}\setminus\{i\}}\,a_j} \leq \frac{1}{n^{n+1}}\,.$$ This is an easy AM-GM exercise.

Further hint, apply AM-GM to the denominator $\sum_{j\in\{0,1,2,\ldots,n\}\setminus\{i\}}\,a_j$ for $i=0,1,2,\ldots,n$.

Prove that $\tfrac{a_1a_2\cdots a_n(1-a_1-a_2-\cdots-a_n)}{(a_1+a_2+\cdots+a_n)(1-a_1)\cdots(1-a_n)} \leq \frac{1}{n^{n+1}}.$

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