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Algebra Difficulty 6.7 National olympiad Prove it

42. α,β,x1,x2,⋯ ,xn(n⩾1)\alpha, \beta, x_{1}, x_{2}, \cdots, x_{n}(n \geqslant 1) are positive numbers, and x1+x2+⋯+xn=1x_{1}+x_{2}+\cdots+x_{n}=1, prove the inequality: x13αx1+βx2+x23αx2+βx3+⋯+xn3αxn+βx1⩾1n(α+β)⋅\frac{x_{1}^{3}}{\alpha x_{1}+\beta x_{2}}+\frac{x_{2}^{3}}{\alpha x_{2}+\beta x_{3}}+\cdots+\frac{x_{n}^{3}}{\alpha x_{n}+\beta x_{1}} \geqslant \frac{1}{n(\alpha+\beta)} \cdot (2002 Moldova National Training Team Problem)

Solution

42. By the generalization of Cauchy's inequality, we have
(x13αx1+βx2+x23αx2+βx3+⋯+xn3αxn+βx1)[(αx1+βx2)+(αx2+βx3)+⋯+(αxn+βx1)](1+1+⋯+1)⩾(x1+x2+⋯+xn)3\begin{array}{l} \left(\frac{x_{1}^{3}}{\alpha x_{1}+\beta x_{2}}+\frac{x_{2}^{3}}{\alpha x_{2}+\beta x_{3}}+\cdots+\frac{x_{n}^{3}}{\alpha x_{n}+\beta x_{1}}\right) \\ {\left[\left(\alpha x_{1}+\beta x_{2}\right)+\left(\alpha x_{2}+\beta x_{3}\right)+\cdots+\left(\alpha x_{n}+\beta x_{1}\right)\right](1+1+\cdots+1) \geqslant} \\ \left(x_{1}+x_{2}+\cdots+x_{n}\right)^{3} \end{array}

Since x1+x2+⋯+xn=1x_{1}+x_{2}+\cdots+x_{n}=1, we have
x13αx1+βx2+x23αx2+βx3+⋯+xn3αxn+βx1⩾1n(α+β)\frac{x_{1}^{3}}{\alpha x_{1}+\beta x_{2}}+\frac{x_{2}^{3}}{\alpha x_{2}+\beta x_{3}}+\cdots+\frac{x_{n}^{3}}{\alpha x_{n}+\beta x_{1}} \geqslant \frac{1}{n(\alpha+\beta)}

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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.