来自人教群的一道简单3元轮换分式不等式

kuing

\begin{align*}
\sum\frac{b^3c}{a^3b+a^2c^2}&=\sum\frac{(b^2c)^2}{a^3b^2c+a^2bc^3} \\
& \geqslant \frac{\left( \sum b^2c \right)^2}{\sum(a^3b^2c+a^2bc^3)} \\
& =\frac{\left( \sum b^2c \right)^2}{abc\sum(a^2b+c^2a)} \\
& =\frac{\sum b^2c}{2abc} \\
& \geqslant \frac32.
\end{align*}