it is possible? 三元分式不等式(x+1)(y+1)(z+2)/(xy+x+y)(z+1)

tommywong

MihaiT:

Dear frend, this is true?
$\displaystyle\sum_{cyc}\frac{x+1}{x}\ge\sum_{cyc}\frac{(x+1)(y+1)(z+{\color{red}2})}{(xy+x+y)(z+1)}$

for any x,y,z>0

thanks very much

kuing

令 `x=1/a-1`, `y=1/b-1`, `z=1/c-1`, `0<a`, `b`, `c<1`，原不等式化为
\[\sum\frac1{1-a}\geqslant\sum\frac{1+c}{1-ab},\]作差有
\[\frac1{1-a}-\frac{1+a}{1-bc}=\frac{a^2-bc}{(1-a)(1-bc)},\]由
\begin{align*}
a^2-bc-(b^2-ca)&=(a-b)(a+b+c),\\
(1-a)(1-bc)-(1-b)(1-ca)&=-(a-b)(1-c),
\end{align*}可知以下两序列为同序
\[\{a^2-bc,b^2-ca,c^2-ab\}, \left\{ \frac1{(1-a)(1-bc)},\frac1{(1-b)(1-ca)},\frac1{(1-c)(1-ab)} \right\},\]故由切比雪夫不等式有
\[\sum\frac{a^2-bc}{(1-a)(1-bc)}\geqslant\frac13\sum(a^2-bc)\sum\frac1{(1-a)(1-bc)}\geqslant0,\]即得证。