安振平老师的4908号不等式问题的证明

题目:设正数$a_{1},a_{2},\cdots,a_{n}$满足$a_{1}+a_{2}+\cdots+a_{n}=1$,求证:

$\left(\frac{1}{a_{1}^{3}}-1\right)\left(\frac{1}{a_{2}^{3}}-1\right)\cdots\left(\frac{1}{a_{n}^{3}}-1\right)\geq (n^{3}-1)^{n}$.

证明:由已知及AM-GM不等式可得:

$\frac{1}{a_{i}^{3}}-1=\frac{1-a_{i}^{3}}{a_{i}^{3}}=\frac{(1-a_{i})(1+a_{i}+a_{i}^{2})}{a_{i}^{3}}=\frac{[(a_{1}+a_{2}+\cdots+a_{n})-a_{i}][(a_{1}+a_{2}+\cdots+a_{n})^{2}+a_{i}\cdot(a_{1}+a_{2}+\cdots+a_{n})+a_{i}^{2}]}{a_{i}^{3}}\geq \frac{n-1}{a_{i}^{3}}\left(\frac{a_{1}a_{2}\cdots a_{n}}{a_{i}}\right)^{\frac{1}{n-1}}\left[n^2(a_{1}a_{2}\cdots a_{n})^{\frac{2}{n}}+na_{i}(a_{1}a_{2}\cdots a_{n})^{\frac{1}{n}}+a_{i}^{2}\right]$

$\geq \frac{n-1}{a_{i}^{3}}\left(\frac{a_{1}a_{2}\cdots a_{n}}{a_{i}}\right)^{\frac{1}{n-1}}\cdot  (n^2+n+1)\left[\left[(a_{1}a_{2}\cdots a_{n})^{\frac{2}{n}}\right]^{n^2}\cdot a_{i}^{n}(a_{1}a_{2}\cdots a_{n})\cdot a_{i}^2\right]^{\frac{1}{n^2+n+1}}$

$=\frac{n^3-1}{a_{i}^3}(a_{1}a_{2}\cdots a_{n})^{\frac{2n+1}{n^2+n+1}+\frac{1}{n-1}}a_{i}^{\frac{n+2}{n^2+n+1}-\frac{1}{n-1}}=\frac{n^3-1}{a_{i}^3}(a_{1}a_{2}\cdots a_{n})^{\frac{3n^2}{n^3-1}}a_{i}^{\frac{-3}{n^3-1}}(i=1,2,\cdots,n)$.

于是

$\left(\frac{1}{a_{1}^{3}}-1\right)\left(\frac{1}{a_{2}^{3}}-1\right)\cdots\left(\frac{1}{a_{n}^{3}}-1\right)\geq \frac{n^3-1}{a_{1}^3}(a_{1}a_{2}\cdots a_{n})^{\frac{3n^2}{n^3-1}}a_{1}^{\frac{-3}{n^3-1}}\cdot \frac{n^3-1}{a_{2}^3}(a_{1}a_{2}\cdots a_{n})^{\frac{3n^2}{n^3-1}}a_{2}^{\frac{-3}{n^3-1}}\cdots \frac{n^3-1}{a_{n}^3}(a_{1}a_{2}\cdots a_{n})^{\frac{3n^2}{n^3-1}}a_{n}^{\frac{-3}{n^3-1}}$

$=(n^3-1)^n\frac{(a_{1}a_{2}\cdots a_{n})^{\frac{3n^3}{n^3-1}-\frac{3}{n^3-1}}}{(a_{1}a_{2}\cdots a_{n})^3}=(n^{3}-1)^{n}$.

故原不等式成立.

posted @ 2019-03-12 08:02 听竹居士的博客 阅读(...) 评论(...) 编辑 收藏