安振平老师的3471号不等式问题的证明

问题 设$a,b,c>0, \frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}=1$,求证: $(a^2-1)(b^2-1)+(b^2-1)(c^2-1)+(c^2-1)(a^2-1)\geq 27$.

证明: 令$\frac{1}{a+1}=x,\frac{1}{b+1}=y,\frac{1}{c+1}=z$,则$a=\frac{1}{x}-1,b=\frac{1}{y}-1,c=\frac{1}{z}-1$且$x+y+z=1(x>0,y>0,z>0)$,从而

原不等式可化为

$(x+y+z)^2\cdot\sum{\frac{(x+y-z)(x-y+z)}{y^2z^2}}\geq 27\Leftrightarrow (x+y+z)^2\cdot \sum{x^2(x+y-z)(x-y+z)}\geq 27x^2y^2z^2$.      (1)

又由Schur不等式可知

$\sum{x^2(x+y-z)(x-y+z)}-xyz(x+y+z)=\sum{x^2(x-y)(x-z)}+\sum{yz(y-z)^2}\geq 0$.

所以

$\sum{x^2(x+y-z)(x-y+z)}\geq xyz(x+y+z)$.                                (2)

由不等式(2)及均值不等式可得

$(x+y+z)^2\cdot\sum{x^2(x+y-z)(x-y+z)}\geq xyz(x+y+z)^3\geq 27x^2y^2z^2$.

即不等式(1)成立,故原不等式获证.

 

posted @ 2017-02-20 08:33 听竹居士的博客 阅读(...) 评论(...) 编辑 收藏