Proving inequality $\prod _{i=1}^n\frac {1-a_i} {a_i}\geqslant \left( n-1\right) ^n$

Question

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 $$\prod _{i=1}^n\dfrac {1-a_i} {a_i}\geqslant \left( n-1\right)^n.$$

Since the most basic version of Lagrange multipliers yields the solution, you really should indicate what you tried. — Did, Aug 06 '12 at 19:31

score 6 · Accepted Answer · answered Aug 06 '12 at 19:50

By AM-GM

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

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.

Proving inequality $\prod _{i=1}^n\frac {1-a_i} {a_i}\geqslant \left( n-1\right) ^n$

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