Please post your solutions. This is just a solution template to write up your solutions in a nice way and formatted in LaTeX. But maybe your solution is so well written that this is not required finally. For more information and instructions regarding the ISL/ILL problems please look here: introduction for the IMO ShortList/LongList project and regardingsolutions

Put $a_{n+1}=1-(a_1+a_2+\ldots+a_n)$. The inequality becomes $\frac{a_1a_2\ldots a_{n+1}}{(1-a_1)(1-a_2)\ldots (1-a_{n+1})}\le \frac 1{n^{n+1}}$, where $a_i$ are positive and $\sum_1^{n+1}a_i=1$. This is a mere application of AM-GM to the denominators $A_i=1-a_i=a_1+a_2+\ldots+a_{i-1}+a_{i+1}+\ldots+a_n+a_{n+1}$. We get $A_i\ge n\sqrt[n]{\frac{\prod a_k}{a_i}}$. After multiplying all of these we get the desired result.

Its too bad that not all inequalities are like this one... Solution Define $a_{n+1}$ such that $\sum^{n+1}_{k=1} a_k = 1$. Cross multiplication and substitution yields that the desired inequality is equivalent to \[n^{n+1} \le \frac{(1-a_1)(1-a_2) \dots (1-a_{n+1})}{a_1 a_2 \dots a_{n+1}} \quad (*)\] By AM-GM we have that \[n\sqrt[n] {\frac{a_1 a_2 \dots a_{n+1}}{a_i}} \le a_1 + a_2 + \dots + a_{i-1} + a_{i+1} + \dots + a_{n+1} = 1-a_i\] Multiplying the above for each $i$ and dividing by $a_1 a_2 \dots a_{n+1}$ yields $(*)$ and the proof is complete.

Proceeding in a manner similar to the above posts, $ n^{n+1}\le\frac{(1-a_1)(1-a_2)\dots (1-a_{n+1})}{a_1 a_2\dots a_{n+1}}$ is easily proven via smoothing with $(a, b) \rightarrow (\frac{a+b}{2}, \frac{a+b}{2})$. The algebra of verifying the smoothing is tedious but straight forward.

$a_1+a_2+...+a_n/n>=( a_1.a_2...a_n)^1/n$. It becomes $1/n^n>= a_1.a_2...a_n/a_1+a_2+...+a_n$ Take is as (1). Now there is an inequality known as weitrass inequality. It will give $1-(a_1+a_2+...+a_n)<=(1-a_1)...(1-a_n)$ Which will become $1-(a_1+a_2+...+a_n)/(1-a_1)...(1-a_n)<=1$ Take it as (2). Multiply (1) and(2).you will.get it less than $1/n^n$ which is all less than $1/n^{n+1}$. Hence shown

More storage 1998 A1 wrote: Let $a_{1},a_{2},\ldots ,a_{n}$ be positive real numbers such that $a_{1}+a_{2}+\cdots +a_{n}<1$. Prove that \[ \frac{a_{1} a_{2} \cdots a_{n} \left[ 1 - (a_{1} + a_{2} + \cdots + a_{n}) \right] }{(a_{1} + a_{2} + \cdots + a_{n})( 1 - a_{1})(1 - a_{2}) \cdots (1 - a_{n})} \leq \frac{1}{ n^{n+1}}. \] Silly Call $1-(a_1+\dots+a_n)$ as $b$ (it really is decoration). Then we just want $(1-a_1) \dots (1-a_n)(1-b) \ge n^{n+1}a_1\dots a_nb$. However, notice that $1-a_i=a_1+\dots+a_{i-1}+a_{i+1}+\dots+a_n+b \geqslant n\sqrt[n]{a_1\dots a_{i-1}a_{i+1}\dots a_nb}$ and multiplying all of them gives the result. $\blacksquare$

Let $a_i=a,a_j=b$ have fixed sum $c<1$. Note \[\frac{ab}{(1-a)(1-b)}\le t\iff ab\le t-tc+tab\iff ab(1-t)\le t(1-c)\]where $t$ is the maximum of the expression. Clearly $t<1$ and so equality would have to be achieved at $a=b$. Thus by a smoothing argument each $a_i$ is equal. Let $\sum_i a_i=c$ so the desired is \[\frac{1}{n^{n+1}}\ge \frac{\frac{c^n}{n^n}[1-c]}{c\cdot (1-\frac cn)^n}=\frac{c^{n-1}[1-c]}{(n-c)^n}\iff (n-c)^n\ge n^{n+1}c^{n-1}[1-c].\]The result is obvious at $n=1$ so we can assume $n>1$. Let $c=\frac{n}{n+1}-\frac{d}{n+1}$ so we need to check \[\left(n^2+d\right)^n \ge n^{n+1}\left(n-d\right)^{n-1}\left(1+d\right)\]and have $-1<d<n$. Since the result is true at the extrema, it suffices to check cases where the derivatives of both sides are equal. In these cases, \[n(n^2+d)^{n-1}=-(n-1)n^{n+1}(n-d)^{n-2}\cdot (1+d)+n^{n+1}(n-d)^{n-1}.\]Equivalently, \[(n^2+d)^{n-1}=n^n(n-d)^{n-2}\cdot [(n-d)-(n-1)(1+d)]=n^n(n-d)^{n-2}\cdot [1-nd].\]Equality can only hold at $d=0$ because $d>0$ increases the left side and decreases the right side whereas $d<0$ increases the right side and decreases the left side. But at $d=0$ the result is obvious, so we are done.

Previous Wrong Solution Write $a_{n+1}=1-a_1-a_2\ldots a_n >0$. Thus, our expression becomes \[\prod_{i=1}^{n+1} \frac{a_i}{1-a_i} = e^{\sum \ln(\frac{a_i}{1-a_i})}\]Note that $\ln(\frac{x}{1-x})$ has a second derivative of $\frac{1}{x(1-x)}$ and is thus convex over $0<x<1$, thus, by jensens \[\sum \ln(\frac{a_i}{1-a_i}) \geq (n+1)\cdot \ln(\frac{1}{n})\]Thus, \[\prod_{i=1}^{n+1} \frac{a_i}{1-a_i}\geq e^{(n+1)\cdot \ln(\frac{1}{n})} = \left(\frac{1}{n}\right)^{n+1}\]and we are done $\blacksquare$. Edit: The second derivative is in fact \[\frac{2x-1}{(x-1)^2x^2}\]New Correct Solution Write $a_{n+1}=1-a_1-a_2\ldots a_n >0$. Now, note that \[\prod_{i=1}^{n+1} \sum_{j\neq i} a_i \leq \prod_{i=1}^{n+1} n\cdot \sqrt[n]{\prod_{j\neq i}a_i} \leq n^{n+1} \cdot \prod a_i\]But wait! this is the denominator of our new expression. The given expression becomes \[\frac{\prod a_i}{\prod (1-a_i)}=\frac{\prod a_i}{\prod (\sum_{j\neq i} a_j)}\leq \frac{\prod_{i=1}^{n+1}}{n^{n+1}\prod_{i=1}^{n+1}}=\frac{1}{n^{n+1}}\]and we are done.

We do the trick of letting $a_{n+1}=1-(a_1+a_2+\ldots+a_n)$. Then we must have $a_1+\ldots+a_{n+1}=1$, and we want to prove that $$\frac{a_1\ldots a_{n+1}}{(1-a_1)\ldots(1-a_{n+1})}\le \frac{1}{n^{n+1}}.$$Now notice that $$1-a_i=(a_1+\ldots+a_{n+1})-a_i\ge n\sqrt[n]{\frac{a_1\ldots a_{n+1}}{a_i}}.$$Applying this to all the terms in the denominator gives us \begin{align*} \frac{a_1\ldots a_{n+1}}{(1-a_1)\ldots(1-a_{n+1})} & \le \frac{a_1\ldots a_{n+1}}{n^{n+1}\sqrt[n]{\frac{a_1\ldots a_{n+1}}{a_1}}\ldots \sqrt[n]{\frac{a_1\ldots a_{n+1}}{a_{n+1}}}} \\ &= \frac{a_1\ldots a_{n+1}}{n^{n+1}\sqrt[n]{a_1^n\ldots a_{n+1}^n}} \\ &= \frac{a_1\ldots a_{n+1}}{n^{n+1}(a_1\ldots a_{n+1})} \\ &= \frac{1}{n^{n+1}}.\end{align*}The proof is complete. Remark: Try stupid stuff first. Don't be like me who takes two hours to solve the problem because I realize that if we take the natural log of both sides, we essentially need to prove that if $f(x)=\ln\left(\frac{x}{1-x}\right)$, we have $$f(a_1)+\ldots+f(a_{n+1})\le (n+1)f\left(\frac{1}{n}\right).$$Then I find $$f''(x)=\frac{2x-1}{(x(x-1))^2}$$so Jensen doesn't work, but I see that there is exactly one inflection point in $(0,1)$, so $n-1$ EV is applicable (you should not need $n-1$ EV on an A1)... You can see how this approach of trying fancy things first fails miserably.

Solution. Define $a_{n + 1} = 1 - (a_1 + \cdots + a_n)$, and note that $a_{n + 1} > 0$ and $a_1 + \cdots + a_{n + 1} = 1$. Then by AM-GM we have \begin{align*} 1 - a_i &= a_1 + \cdots + a_{i - 1} + a_{i + 1} + \cdots + a_{n + 1} \\ &\geq n(a_1\cdots a_{i - 1} a_{i + 1} \cdots a_{n + 1})^{\frac{1}{n}}. \end{align*}Taking the product over $i$, $$ \prod_i (1 - a_i) \geq n^{n + 1} a_1 \cdots a_{n + 1}, $$that is, $$ \frac{a_1 \cdots a_n (1 - (a_1 + a_2 + \cdots + a_n))}{(a_1 + \cdots + a_n)(1 - a_1) \cdots (1 - a_n)} \leq \frac{1}{n^{n + 1}}. $$

So, by GM-HM, $$\sqrt[n]{\left(\frac{1}{a_1}-1\right) \left(\frac{1}{a_2}-1\right)\ldots \left(\frac{1}{a_n}-1\right)}\geq \frac{n}{\frac{a_1}{1-a_1}+\frac{a_2}{1-a_2}+\ldots+\frac{a_n}{1-a_n}}.$$Now it is suffices to show that $$\left(\frac{1}{\frac{a_1}{1-a_1}+\frac{a_2}{1-a_2}+\ldots+\frac{a_n}{1-a_n}}\right)^n\geq n\left(\frac{1}{a_1+a_2+\ldots+a_n}-1\right).$$ By Jensen's inequality, $$\frac{a_1}{1-a_1}+\frac{a_2}{1-a_2}+\ldots+\frac{a_n}{1-a_n}\leq n\frac{\frac{a_1+a_2+\ldots+a_n}{n}}{1-\frac{a_1+a_2+\ldots+a_n}{n}}=\frac{a_1+a_2+\ldots+a_n}{1-\frac{a_1+a_2+\ldots+a_n}{n}}=\frac{n(a_1+a_2+\ldots+a_n)}{n-(a_1+a_2+\ldots+a_n)}.$$Therefore it is suffices to show that $$\left(\frac{1}{a_1+a_2+\ldots+a_n}-\frac{1}{n}\right)^{n} \geq n\left(\frac{1}{a_1+a_2+\ldots+a_n}-1\right).$$Let $t=\frac{1}{a_1+a_2+\ldots+a_n}>1$. We need \[ \left(t-\frac{1}{n}\right)^{n} \geq n\left(t-1\right).\]Define $f:\mathbb{R}_{>1}\to \mathbb{R}$, such that $f(t)=\left(t-\frac{1}{n}\right)^{n} - n\left(t-1\right)$. We want to find the minimum of $f$. We have $f'(t)=n\left(t-\frac{1}{n}\right)^{n-1}-n$ and $f''(t)=n(n-1)\left(t-\frac{1}{n}\right)^{n-2}>0 \forall t>1$. Also note that $f'(t)=0\Longleftrightarrow t=\frac{n+1}{n}$ and thus the minimum of the function $f$ is $$f\left(\frac{n+1}{n}\right)=\left(\frac{n+1}{n}-\frac{1}{n}\right)^{n} - n\left(\frac{n+1}{n}-1\right)=1-1=0.$$Hence, we have proven the inequality \[ \left(t-\frac{1}{n}\right)^{n} \geq n\left(t-1\right),\]where the equality holds iff $t=\frac{n+1}{n}$. We conclude that that the equality in our original inequality holds iff $a_1=a_2=\ldots=a_n=\frac{1}{n+1}$. We are done.

It is equivalent to showing that $\left(\frac{1}{a_1}-1\right)\left(\frac{1}{a_2}-1\right)\dots\left(\frac{1}{a_{n+1}}-1\right)\geq n^{n+1}$ for positive reals $a_1,a_2,\dots,a_{n+1}$ with $n\geq 1$ that sum to $1$ (in the problem, $a_{n+1}$ is replaced with $1-(a_1+a_2+\dots+a_n)$). Then $$\frac{1}{a_1}-1=\frac{a_2+\dots+a_{n+1}}{a_1}\geq \frac{n\sqrt[n]{a_2a_3\dots a_{n+1}}}{a_1}$$and multiplying together gives the desired result. Equality holds when $a_1=a_2=a_3=\dots=a_{n+1}=\frac{1}{n+1}$.

For $k\leq n$, let $b_n=a_n$, and let $b_{n+1}=1-\sum_{k=0}^n a_k$. Note $$\sum_{k=1}^{n+1} b_k=1$$and all $b_k$ are positive. By AM-GM, for all $1\geq k\leq n+1$, $$\sum_{j\neq k}b_k\geq n\sqrt[n]{\prod_{j\neq k}b_k}$$So, $$\prod_{k=1}^{n+1} \sum_{j\neq k}b_j\geq \prod_{k=1}^{n+1} n\sqrt[n]{\prod_{j\neq k}b_j}$$$$\prod_{k=1}^{n+1} \sum_{j\neq k}b_j\geq n^{n+1}\prod_{k=1}^{n+1} \sqrt[n]{\prod_{j\neq k}b_j}$$$$\prod_{k=1}^{n+1} 1-b_k\geq n^{n+1}\prod_{k=1}^{n+1} b_k$$$$\frac{\prod_{k=1}^{n+1} 1-b_k}{\prod_{k=1}^{n+1} b_k}\geq n^{n+1}$$$$\frac{\prod_{k=1}^{n+1} b_k}{\prod_{k=1}^{n+1} 1-b_k}\leq \frac{1}{n^{n+1}}$$$$\frac{\prod_{k=1}^{n+1} b_k}{\prod_{k=1}^{n+1} 1-b_k}\leq \frac{1}{n^{n+1}}$$$$\frac{a_{1} a_{2} \cdots a_{n} \left[ 1 - (a_{1} + a_{2} + \cdots + a_{n}) \right] }{(a_{1} + a_{2} + \cdots + a_{n})( 1 - a_{1})(1 - a_{2}) \cdots (1 - a_{n})} \leq \frac{1}{ n^{n+1}}$$

ISL marabot solve Let $1-(a_1+a_2+\ldots+a_n)=a_{n+1}$ and take the reciprocal to get that the original inequality is equivalent to \[\left(\frac{1-a_1}{a_1}\right)\cdot \left(\frac{1-a_2}{a_2}\right)\cdots \left(\frac{1-a_{n+1}}{a_{n+1}}\right)\ge n^{n+1}.\] Now we have $1-a_i=\sum_{k=1, k\ne i}^{n+1} a_k\ge n\sqrt[n]{\prod_{k=1, k\ne i}^{n+1} a_k}$. So the numerator is greater than or equal to $n^{n+1}a_1\cdot a_2\cdots a_{n+1}$. Dividing by the denominator gives the desired result.

First, throw out the $n = 1$ case and define $a_0 = 1 - \sum\limits_{i=1}^{n} a_i$. Then, we have $\sum\limits_{i=1}^{n} a_i = 1$, and we want to show that \[ \prod\limits_{i=0}^{n} \frac{a_i}{1-a_i} \leq \frac{1}{n^{n+1}} \]and notice that equality holds when $a_0 = a_1 = \cdots = a_n = \frac{1}{n+1}$. Now, we smooth. Say that there is some set of values $(a_0,a_1,\ldots,a_n)$ such that the value of the product is maximal, and assume for the sake of contradiction that there exist some $a_i \neq a_j$. Then, we are done if \[ \frac{a_i}{1-a_i}\frac{a_j}{1-a_j} \leq \left(\frac{a_i+a_j}{2-a_i-a_j}\right)^2 \]as that would imply that there was some larger value of the product by taking the values \[ \left(a_0,a_1,\ldots,a_{i-1},\frac{a_i+a_j}{2},a_{i+1},\ldots,a_{j-1},\frac{a_i+a_j}{2},a_{j+1},\ldots,a_n\right) \]Now, we prove that this set is actually larger. We use the subsitution $a_i + a_j = 2u$ and $a_i - a_j = 2v$. Then, we can rewrite the equation we want to prove as \[ \frac{a_i}{1-a_i}\frac{a_j}{1-a_j} \leq \left(\frac{a_i+a_j}{2-a_i-a_j}\right)^2 \Rightarrow \frac{u^2 - v^2}{(1-u)^2 - v^2} < \frac{u^2}{(1-u)^2} \]after some algebra, which further rearranges into \[ \frac{u^2 - (1-u)^2}{(1-u)^2 - v^2} < \frac{u^2 - (1-u)^2}{(1-u)^2} \]which is true because $2u = a_i + a_j < 1$.

grobber wrote: Put $a_{n+1}=1-(a_1+a_2+\ldots+a_n)$. The inequality becomes $\frac{a_1a_2\ldots a_{n+1}}{(1-a_1)(1-a_2)\ldots (1-a_{n+1})}\le \frac 1{n^{n+1}}$, where $a_i$ are positive and $\sum_1^{n+1}a_i=1$. This is a mere application of AM-GM to the denominators $A_i=1-a_i=a_1+a_2+\ldots+a_{i-1}+a_{i+1}+\ldots+a_n+a_{n+1}$. We get $A_i\ge n\sqrt[n]{\frac{\prod a_k}{a_i}}$. After multiplying all of these we get the desired result. Very nice idea

Let $a_{n+1}=1-(a_1+a_2+ \cdots a_n)$. We have that the inequality becomes $\left( \frac{1}{a_1}-1 \right) \left( \frac{1}{a_2}-1 \right) \cdots \left( \frac{1}{a_{n+1}}-1 \right) \ge n^{n+1}$. We note that $$\frac{1}{a_1}-1 = \frac{a_2+a_3 + \cdots + a_{n+1}}{a_1} \ge \frac{n \sqrt[n]{a_2a_3 \cdots a_{n+1}}}{a_1}$$and multiplying similar inequalities gives the result.

Let $a_{n+1}$ be the positive real number such that $a_{1}+a_{2}+\cdots +a_{n+1}=1.$ The desired inequality becomes \[ (na_{1}) (na_{2}) \cdots (na_{n}) (na_{n+1}) \le ( 1 - a_{1})(1 - a_{2}) \cdots (1 - a_{n})(1-a_{n+1}) \]By AM-GM we have \[1-a_{i}=a_1+a_2+\dots+a_{n+1}-a_i\ge n\cdot \sqrt[n]{\frac{a_1a_2\cdots a_{n+1}}{a_i}}\]Multiplying the analogous gives the result.

Define $1-(a_1+a_2+\dots+a_n)=a_{n+1}$; the LHS is equal to \[\frac{a_1a_2\dots a_{n+1}}{\prod_{\text{cyc}}(a_1+a_2+\dots+a_n)}\le \frac{a_1a_2\dots a_{n+1}}{\prod_{\text{cyc}}(n\sqrt[n]{a_1a_2\dots a_n})}=\frac{1}{n^{n+1}}\]so we are done. $\blacksquare$

ISL Marabot solve Let, $a_{n+1} = 1-\sum\limits_{i=1}^na_i$. So, $\sum\limits_{i=1}^{n+1}a_i =1$. Now for each $1\leq k \leq n+1$, \begin{align*} \sqrt[n]{\prod_{i\neq k}a_i} &\leq \dfrac{\sum_{i\neq k}a_i}n\\ \implies \sqrt[n]{\prod_{i\neq k}a_i} &\leq \dfrac{(1-a_k)}n \end{align*}And, multiplying all of these gives, \begin{align*} \prod_{i=1}^{n+1}a_i &\leq \dfrac{\prod\limits_{i=1}^{n+1} (1-a_i)}{n^{n+1}}\\ \implies \dfrac{\prod\limits_{i=1}^{n+1}a_i}{\prod\limits_{i=1}^{n+1} (1-a_i)} &\leq \dfrac1{n^{n+1}} \\ \end{align*}Which is what we wanted. $\blacksquare$

Let $a=a_1+a_2+\dots+a_n$. Taking $\ln$ of both sides, we can first optimize $$\sum_{\text{cyc}}\ln\left(\frac{a_1}{1-a_1}\right).$$Using log properties, we find that the second derivative of the term we are summing is $$-\frac{1}{x^2}+\frac{1}{(1-x)^2}$$which is zero at $x=\frac12$, so there is exactly one inflection point, meaning that we can use $n-1$ EV to have WLOG $a_1=a_2=\dots=a_{n-1}$. Our wanted inequality is now $$\frac{a_1^{n-1}\cdot a_n\cdot(1-a)}{a\cdot(1-a_1)^{n-1}\cdot(1-a_n)}\le\frac{1}{n^{n+1}}.$$By AM-GM, we can instead just show that $$(n-1)\cdot\frac{a_1}{1-a_1}+\frac{a_n}{1-a_n}+\frac{1-a}{a}\le\frac{n+1}{n}.$$Adding $n+1$ to both sides, we want $$\frac{n-1}{1-a_1}+\frac{1}{1-a_n}+\frac{1}{(n-1)a_1+a_n}\le\frac{(n+1)^2}{n}.$$We can resort to differentiating the LHS with respect to $a_n$ to get $$\frac{1}{(1-a_n)^2}-\frac{1}{((n-1)a_1+a_n)^2}$$which is zero when $a_n=\frac{1-a_1(n-1)}{2}$. The second derivative is positive for the whole interval, so that root is the global minimum for the interval. Plugging this back in, we want to show that $$\frac{n-1}{1-a_1}+\frac{4}{1+a_1(n-1)}\le\frac{(n+1)^2}{n}.$$We can do this using another direct differentiation with respect to $a_1$. We get $$\frac{n-1}{\left(1-a_1\right)^2}-\frac{4\left(n-1\right)}{\left(\left(n-1\right)a_1+1\right)^2}$$and we have to check $$a_1=0,\frac1{n+1},\frac1{n-1}.$$It is easy to check that the wanted inequality does hold true for all of these, so we are done.

Let $a_{n+1}=1-\sum_{i=1}^n a_i$, so $\sum_{i=1}^{n+1} a_i = 1$. It then suffices to show $$\prod_{i=1}^{n+1} \left(\frac{1-a_i}{a_i}\right) \geq n^{n+1},$$which is clearly true by AM-GM on the numerator. $\blacksquare$

This abomination of a solution is caused by the fact that I did not see the clean way.

Let $a_{n+1} = 1-a_1-a_2-\cdots-a_n$. Then, the inequality is equivalent to $$\frac{a_1a_2\cdots a_{n+1}}{(1-a_1)(1-a_2)\cdots (1-a_{n+1})} \leq \frac 1{n^{n+1}}.$$However, observe that $$1-a_1 = a_2+a_3+\cdots+a_{n+1} \geq n\sqrt[n]{a_2a_3\cdots a_{n+1}},$$so multiplying this inequality cyclically in the denominator yields the result.

Let $a_{n+1} = 1- \sum a_i$. Then it remains to show $$\prod \frac{a_i}{1-a_i} \leq \frac 1{n^{n+1}}.$$This is easy with AM-GM, as $$\prod \frac{a_i}{1-a_i} \leq \prod \frac{a_i^{(n+1)/n}}{n \cdot \prod a_i} = \frac 1{n^{n+1}}.$$

Let $a_{n+1}=1-\sum_{i=1}^na_i$, thus the inequality is equal to: $\frac{\prod_{i=1}^{n+1}a_i}{(a_1+... a_n)\cdots (a_1+...a_{n+1})}$ Furthermore notice that:\begin{align*} a_1+\dots +a_n\ge n\sqrt[n]{\prod_{i=1}^n a_i}\end{align*}$ $ \begin{align*}\vdots\end{align*}\begin{align*}a_1+\dots +a_{n-1}+a_{n+1}\ge n\sqrt[n]{a_1\cdots a_{n-1}a_{n+1}} \end{align*}By multiplying the inequalities: $LHS\ge n^{n+1}\prod_{i=1}^{n+1}a_i$ Thus $LHS \le\frac{\prod_{i=1}^{n+1}a_{n}}{n^{n+1}\prod_{i=1}^{n+1}a_n}=\frac{1}{n^{n+1}}$. QED And we are done!

Let $a_{n+1} = 1-\sum_{i=1}^n a_i$. We claim the maximum can be achieved when all the variables are equal. Assume not. WLOG, $a_1 \neq a_2$. Let $x=a_1, y=a_2$. But consider \[ \frac{\left( \frac{x+y}2 \right)^2}{\left(1 - \frac{x+y}2 \right)^2} - \frac{xy}{(1-x)(1-y)} = \frac{(x-y)^2 (1-x-y)}{(1-x)(1-y)(2-x-y)^2} \ge 0, \]so making them equal (with the same sum) results in a product at least as big as the original one. The conclusion follows.

orl wrote: Let $a_{1},a_{2},\ldots ,a_{n}$ be positive real numbers such that $a_{1}+a_{2}+\cdots +a_{n}<1$. Prove that \[ \frac{a_{1} a_{2} \cdots a_{n} \left[ 1 - (a_{1} + a_{2} + \cdots + a_{n}) \right] }{(a_{1} + a_{2} + \cdots + a_{n})( 1 - a_{1})(1 - a_{2}) \cdots (1 - a_{n})} \leq \frac{1}{ n^{n+1}}. \] Putting $a_{n+1}=1-\sum_{k=1}^{n}a_k$, it suffices to prove that $\prod_{i=1}^{n}\frac{(1-a_i)}{a_i}\ge n^{n+1}$. Note that 1-a_i=the sum of all a_j's from 1 to n excluding a_i. Then this inequality follows immediately from $\prod_{i=1}^{n}\frac{(1-a_i)}{a_i}=\frac{\prod_{i}(\sum_{j\ne i}a_j)}{\prod_{i}a_i}\ge \frac{(n-1)^n\prod_{i}(\prod_{j\ne i}a_j)^{\frac{1}{n-1}})}{\prod_{i}a_i}=(n-1)^n$, where the greater or equal to is just the AM-GM. Each of these steps are reversible, so we're done. $\blacksquare$

@above, I think the product should go from $i=1$ to $i=n+1$, no? And then you would get $\prod_{i=1}^{n+1} \geq n^{n+1}$ instead of $\prod_{i=1}^{n}\frac{1-a_i}{a_i} \geq (n-1)^n$.

Sorry, yeah. The Latexing took a long time so I couldn't notice what was wrong. The index would be changed over by 1 (1->n+1) and like you said.

Let $a_{n+1}=1-a_1-a_2-...-a_n$. $\frac{a_1a_2...a_{n+1}}{(1-a_1)(1-a_2)...(1-a_{n+1})}=\Pi_{k=1}^{n+1} \frac{a_k}{1-a_k}$. Note that for $x \neq y$, $\frac{(\frac{x+y}{2})^2}{(1-\frac{x+y}{2})(1-\frac{x+y}{2})} < \frac{xy}{(1-x)(1-y)}$, so turning $x$ and $y$ into their averages makes the product smaller. So all the numbers being equal is the minimum case, which results in $\frac{1}{n^{n+1}}$.

Replace $\sum a_i$ with $a_{n + 1}$, now we have the better condition $a_1 + \cdots a_{n} = 1$, prove $\prod \frac{a_i}{1 - a_i} \le \frac{1}{n-1^n}$. We proceed by smoothing, we show replacing any two $a_i$ with their average does not decrease the product. We desire $\frac{x}{1 - x}\frac{y}{1 - y} \le (\frac{\frac{x + y}{2}}{ 1 - \frac{x + y}{2}})^2 $, equivalently $xy(2 - (x + y))^2 \le (1 - x)(1 - y)(x + y)^2$, standard expansion gives the desired as $xy((x + y)^2 - 4(x + y) + 4) \le (xy - x - y + 1)(x + y)^2$, cancellation gives $(x + y)^3 + 4xy \le 4(x + y)xy + (x + y)^2$. We can write this as $x^3 + y^3 \le (x-y)^2 + xy^2 + yx^2$. We can now write the left hand side as $(x + y)(x - y)^2 + xy^2 + yx^2$, since $x + y \le 1$, the bound is now obvious. Thus applying this operation to the largest and smallest elements of $a_i$ repeatedly, we can see $a_i$ approaches $\frac 1n$. Assume there exists some tuple $b_i$ for which the product is a factor of $x$ greater than the claimed maximum. After applying the operation some arbitrarily large number of times, we can eventually reach a tuple $c_i$ for which each element is a factor $y$ away from $\frac 1n$ for arbitrarily small $y$, so we can then bound the product for $c_i$ as $z$ away from desired for arbitrarily small $z$, choosing $z < x$ forces a contradiction.

Note that there exists a number $a_{n+1}\in (0,1)$ such that $a_1+a_2+\dots+a_n+a_{n+1}=1$. Then we need to prove the nicer inequality $$\frac{a_1a_2\cdots a_{n+1}}{(1-a_1)(1-a_2)\cdots (1-a_{n+1})} \leq \frac{1}{n^{n+1}}.$$But this is the same as proving \begin{align*} & \frac{a_2+a_3+\dots+a_{n+1}}n \cdot \frac{a_1+a_3+\dots+a_{n+1}}n \cdot \frac{a_1+a_2+a_4+\dots+a_{n+1}}n \cdots \frac{a_1+a_2+\dots+a_{n-1}+a_{n+1}}n \cdot \frac{a_1+a_2+\dots+a_{n-1}+a_n}n \\ & \qquad \geq \sqrt[n]{a_2a_3 \cdots a_{n+1}}\sqrt[n]{a_1a_3 \cdots a_{n+1}}\sqrt[n]{a_1a_2 a_4 \cdots a_{n+1}}\cdots \sqrt[n]{a_1a_2\cdots a_{n-1}a_{n+1}}\sqrt[n]{a_1a_2 \cdots a_{n-1}a_n}\end{align*}which is just AM-GM.

Let $a_0 = 1-(a_1+a_2+\cdots + a_n) > 0.$ Then the inequality is equivalent to $$n^{n+1} \leq \prod_{i=0}^{n} \frac{a_0+a_1+\cdots + a_n - a_i}{ a_i}$$which is true by AM-GM. QED