安振平老师的5896号不等式问题的证明

题目：已知$a,b,c>0$,求证：$\frac{a}{1+2b^2c}+\frac{b}{1+2c^2a}+\frac{c}{1+2a^2b}\geq \frac{a+b+c}{1+2abc}.$

证明:由柯西不等式可得

$\frac{a}{1+2b^2c}+\frac{b}{1+2c^2a}+\frac{c}{1+2a^2b}=\frac{a^2}{a+2ab^2c}+\frac{b^2}{b+2abc^2}+\frac{c^2}{c+2a^2bc}$

$\geq \frac{(a+b+c)^2}{(a+b+c)(1+2abc)}=\frac{a+b+c}{1+2abc}.$