IstekOlympiadTeam wrote: Let $n$ be a positive integer and let $a_1,\cdots ,a_n$ be real numbers satisfying $0\le a_i\le 1$ for $i=1,\cdots ,n.$ Prove the inequality \[(1-{a_i}^2)(1-{a_2}^2)\cdots (1-{a_n}^n)\le (1-a_1a_2\cdots a_n)^n.\] I am sure there is some typo. Could you please correct it ?

utkarshgupta wrote: IstekOlympiadTeam wrote: Let $n$ be a positive integer and let $a_1,\cdots ,a_n$ be real numbers satisfying $0\le a_i\le 1$ for $i=1,\cdots ,n.$ Prove the inequality \[(1-{a_i}^2)(1-{a_2}^2)\cdots (1-{a_n}^n)\le (1-a_1a_2\cdots a_n)^n.\] I am sure there is some typo. Could you please correct it ? http://bw15.math.su.se/wp-content/uploads/2015/11/Baltic-Way-2015.pdf

Link shows there is a typo, evth is n-th power instead of 2 on lefthand side of inequality

Note that if any of the $a_i$ is equal to 0 or 1, the inequality is obviously true. Now, assume $0<a_i<1$ for all $i$. Denote $x_i=\ln(a_i^n)$ and $f(x)=\ln(1-e^x)$. Note that $x_i<0$ and hence $f(x_i)$ is well-defined (and real). Then the inequality is equivalent to proving \[f(x_1)+f(x_2)+\dotsc+f(x_n) \le n \cdot f \left( \frac{x_1+x_2+\dotsc+x_n}{n} \right)\]for $x_i<0$. But this is trivial by Jensen since $f''(x)=-\frac{e^x}{(1-e^x)^2}<0$ i.e. $f$ is concave.

AM-GM $ (1-{a_i}^2)(1-{a_2}^2)\cdots (1-{a_n}^n) \le ((1-a_1^n+1-a_2^n+...+a_n^n)/n)^n$ $((1-a_1^n+1-a_2^n+...+a_n^n)/n)^n=(1-(a_1^n+a_2^n+...+a_n^n)/n)^n$ AM-GM again $ a_1a_2..a_n\le (a_1^n+a_2^n+...+a_n^n)/n$ hence, we're finished $((1-a_1^n+1-a_2^n+...+a_n^n)/n)^n\le (1-a_1a_2\cdots a_n)^n.$ => $ (1-{a_i}^n)(1-{a_2}^n)\cdots (1-{a_n}^n)\le (1-a_1a_2\cdots a_n)^n. $

Is there any way to proceed with this idea? Consider 1-coins, 2-coins, ...$n$-coins where the probability an $i$-coin is heads is $a_i$. Consider flipping $n$ 1-coins, $n$ 2-coins, ..., $n$ $n$-coins ($n$ rounds) and mark the results in the table: _______________Coin______1____ 2____ ..._________$n-1$________$n$ Round 1 2 : : $n-1$ $n$ LHS equals the probability that no column has all heads. RHS equals the probability that no row has all heads.

I have just realized that this one is my old inequality, actually the easy side of my old inequality. According to http://bw15.math.su.se/ the competition was held between the 5th and 9th of November, 2015. But I already published it in https://ilkercancicek.wordpress.com/ as Problem 7 in the 24th of September, 2015. https://ilkercancicek.wordpress.com/ is also my website, where I publish the olympiad problems I create. To view my own problem statement and solution please see here: https://ilkercancicek.files.wordpress.com/2015/09/problem-7-e29480-c3a7c3b6zc3bcm.pdf

Easy using mixing variables Fix $a_1 , a_2 , ... , a_{n-2}$ and also fix $a_{n-1}a_n$ now its easy to see that the right hand side is fixed , and in left hand side $(1-{a_1}^n)(...)(1-{a_{n-2}}^n)$ is fixed and now we’ll show that as $a_{n-1}$ and $a_n$ get closer to each other the left hand side gets bigger and bigger , so it just suffices to show that $(1-{a_{n-1}}^n)(1-{a_n}^n) \le (1-{\sqrt{a_{n-1}a_n}}^n)^2$ which is equivalent to $2{\sqrt{a_{n-1}a_n}}^n \le {a_{n-1}}^n + {a_n}^n$ which is trivial by $\textrm{AM-GM}$ so by what we said the maximum of the expression in the left hand hand side happens when all the $a_{i}$ are equal and in this case $\textrm{LHS=RHS}$ so the problem is solved