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Let $a_1,a_2,\ldots ,a_n\in \left( 0,1\right)$ be real numbers such that $\sum\limits_{i=1}^n a_i=1$. Prove that $$\prod _{i=1}^n\dfrac {1-a_i} {a_i}\geqslant \left( n-1\right)^n.$$

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Since the most basic version of Lagrange multipliers yields the solution, you really should indicate what you tried. – Did Aug 6 '12 at 19:31
-I am looking for an elementary proof – Maria Mikolayevskaya Aug 6 '12 at 19:44
Funny that you do not even pretend to answer my question. – Did Aug 6 '12 at 21:16

1 Answer 1

up vote 6 down vote accepted


$$\dfrac {1-a_i} {a_i} = \frac{a_1+a_2+..+a_{i-1}+a_{i+1}+..+a_n}{a_i} \geq (n-1) \frac{\sqrt[n-1]{a_1a_2..a_{i-1}a_{i+1}a_n}}{a_i}$$

Multiply them together and observe that everything in $\prod_{i=1}^n \frac{\sqrt[n-1]{a_1a_2..a_{i-1}a_{i+1}a_n}}{a_i}$ cancels.

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