Inequality of Arithmetic and Geometric Means (AM-GM) states that for all positive real numbers $x_1$, $x_2$, … , $x_n$ it holds that
$$ \frac{ x_1+x_2+…+x_n }{n} \geq \sqrt[n]{ x_1x_2…x_n } $$
The equality holds if there exists $k$, such that $x_i=k$ for all $i=1,2,…,n$.
Problem (IMO Shortlist, 1998)
Let $a_1$, $a_2$, … , $a_n$ be positive numbers with $a_1 + a_2 + … + a_n < 1$. Prove that
$$ \frac{a_1 a_2 … a_n \left( 1 – \left( a_1 + a_2+ … + a_n \right) \right) }{ \left( a_1 + a_2+ … + a_n \right)(1-a_1)(1-a_2) … (1-a_n) } \leq \frac{1}{n^{n+1}} $$
Solution
Let $ a_{n+1} = 1 – \left( a_1 + a_2+ … + a_n \right) $ and $p=a_1 a_2 … a_n a_{n+1}$. Therefore $ a_1 + a_2+ … + a_n = 1 -a_{n+1} $ and the inequality becomes
$$ \frac{a_1 a_2 … a_n a_{n+1} }{(1-a_1)(1-a_2) … (1-a_n)(1-a_{n+1}) } \leq \frac{1}{n^{n+1}} $$
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Learn moreor equivalently
$$ (1-a_1)(1-a_2) … (1-a_n)(1-a_{n+1}) \geq n^{n+1} a_1 a_2 … a_n a_{n+1} $$
By AM-GM Inequality
$$ 1-a_i = a_{1} + a_{2} + … + a_{i-1} + a_{i+1} + … + a_{n} + a_{n+1} \geq n \sqrt[n]{\frac{p}{a_i}} $$
and the needed inequality is the multiplication of these inequalities.