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Algebra Difficulty 6.9 National olympiad Prove it

Example 2 Let ai>0(i=1,2,⋯ ,n),∑i=1nai=1a_{i}>0(i=1,2, \cdots, n), \sum_{i=1}^{n} a_{i}=1, prove: (1a1−1)(1a2−1)⋯(1an−1)⩾(n−1)n\left(\frac{1}{a_{1}}-1\right)\left(\frac{1}{a_{2}}-1\right) \cdots\left(\frac{1}{a_{n}}-1\right) \geqslant(n-1)^{n}.

Solution

Prove ∵1−a1=a2+a3+⋯+an\because 1-a_{1}=a_{2}+a_{3}+\cdots+a_{n}
⩾(n−1)(a2a3⋯an)1n−1\geqslant(n-1)\left(a_{2} a_{3} \cdots a_{n}\right)^{\frac{1}{n-1}}

Similarly, 1−a2⩾(n−1)(a1a3⋯an)1n−1,⋯1-a_{2} \geqslant(n-1)\left(a_{1} a_{3} \cdots a_{n}\right)^{\frac{1}{n-1}}, \cdots,
1−an⩾(n−1)(a1a2⋯an−1)1n−11-a_{n} \geqslant(n-1)\left(a_{1} a_{2} \cdots a_{n-1}\right)^{\frac{1}{n-1}}

Multiplying these nn "component inequalities", we get
(1−a1)(1−a2)⋯(1−an)⩾(n−1)n(a1a2⋯an), thus (1a1−1)(1a2−1)⋯(1an−1)=(1−a1)(1−a2)⋯(1−an)a1a2⋯an⩾(n−1)n.\begin{aligned} & \left(1-a_{1}\right)\left(1-a_{2}\right) \cdots\left(1-a_{n}\right) \\ \geqslant & (n-1)^{n}\left(a_{1} a_{2} \cdots a_{n}\right), \\ & \text { thus }\left(\frac{1}{a_{1}}-1\right)\left(\frac{1}{a_{2}}-1\right) \cdots\left(\frac{1}{a_{n}}-1\right) \\ = & \frac{\left(1-a_{1}\right)\left(1-a_{2}\right) \cdots\left(1-a_{n}\right)}{a_{1} a_{2} \cdots a_{n}} \geqslant(n-1)^{n} . \end{aligned}

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.