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Algebra Difficulty 6.7 National olympiad Prove it

Example 3 If xi>0(i=1,2,⋯ ,n),n⩾2x_{i}>0(i=1,2, \cdots, n), n \geqslant 2, and 11+x1+11+x2+⋯+11+xn=\frac{1}{1+x_{1}}+\frac{1}{1+x_{2}}+\cdots+\frac{1}{1+x_{n}}=
1, prove that: x1x2⋯xn⩾(n−1)nx_{1} x_{2} \cdots x_{n} \geqslant(n-1)^{n}

Solutions — 2

Solution 1

 Proof: Let 11+x1=a1a1+a2+⋯+an,11+x2=a2a1+a2+⋯+an,⋯ ,11+xn=ana1+a2+⋯+an, where a1,a2,⋯ ,an are positive numbers, then x1=a2+a3+⋯+ana1⩾(n−1)a2a3⋯ann−1a1x2=a1+a3+⋯+ana2⩾(n−1)a1a3ana2xn=a1+a2+⋯+an−1an(n−1)a1a2⋯an−1n−1an\begin{array}{l} \text { Proof: Let } \frac{1}{1+x_{1}}=\frac{a_{1}}{a_{1}+a_{2}+\cdots+a_{n}}, \frac{1}{1+x_{2}}=\frac{a_{2}}{a_{1}+a_{2}+\cdots+a_{n}}, \cdots, \\ \frac{1}{1+x_{n}}=\frac{a_{n}}{a_{1}+a_{2}+\cdots+a_{n}}, \text { where } a_{1}, a_{2}, \cdots, a_{n} \text { are positive numbers, then } \\ x_{1}=\frac{a_{2}+a_{3}+\cdots+a_{n}}{a_{1}} \geqslant(n-1) \frac{\sqrt[n-1]{a_{2} a_{3} \cdots a_{n}}}{a_{1}} \\ x_{2}=\frac{a_{1}+a_{3}+\cdots+a_{n}}{a_{2}} \geqslant(n-1) \frac{\sqrt{a_{1} a_{3}} a_{n}}{a_{2}} \\ x_{n}=\frac{a_{1}+a_{2}+\cdots+a_{n-1}}{a_{n}}(n-1) \frac{\sqrt[n-1]{a_{1} a_{2} \cdots a_{n-1}}}{a_{n}} \end{array}

Multiplying the above n−1n-1 inequalities, we get x1x2⋯xn⩾(n−1)nx_{1} x_{2} \cdots x_{n} \geqslant(n-1)^{n}.

Solution 2

Proof: Let 11+x1=a1a1+a2+⋯+an,11+x2=a2a1+a2+⋯+an\frac{1}{1+x_{1}}=\frac{a_{1}}{a_{1}+a_{2}+\cdots+a_{n}}, \frac{1}{1+x_{2}}=\frac{a_{2}}{a_{1}+a_{2}+\cdots+a_{n}}, ⋯ ,11+xn=ana1+a2+⋯+an\cdots, \frac{1}{1+x_{n}}=\frac{a_{n}}{a_{1}+a_{2}+\cdots+a_{n}}, where a1,a2,⋯ ,ana_{1}, a_{2}, \cdots, a_{n} are positive numbers, then
x1=a2+a3+⋯+ana1⩾(n−1)a2a3⋯ann−1a1,x2=a1+a3+⋯+ana2⩾(n−1)a1a3⋯ann−1a2,⋯xn=a1+a2+⋯+an−1an⩾(n−1)a1a2⋯an−1n−1an.\begin{array}{c} x_{1}=\frac{a_{2}+a_{3}+\cdots+a_{n}}{a_{1}} \geqslant(n-1) \frac{\sqrt[n-1]{a_{2} a_{3} \cdots a_{n}}}{a_{1}}, \\ x_{2}=\frac{a_{1}+a_{3}+\cdots+a_{n}}{a_{2}} \geqslant(n-1) \frac{\sqrt[n-1]{a_{1} a_{3} \cdots a_{n}}}{a_{2}}, \\ \cdots \\ x_{n}=\frac{a_{1}+a_{2}+\cdots+a_{n-1}}{a_{n}} \geqslant(n-1) \frac{\sqrt[n-1]{a_{1} a_{2} \cdots a_{n-1}}}{a_{n}} . \end{array}

Multiplying the above n−1n-1 inequalities, we get
x1x2⋯xn⩾(n−1)n.x_{1} x_{2} \cdots x_{n} \geqslant(n-1)^{n} .

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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.