Lời giải:
$P=\frac{a}{b(b^2+a)}+\frac{b}{c(c^2+b)}+\frac{c}{a(a^2+c)}$
$=\frac{1}{b}-\frac{b}{a+b^2}+\frac{1}{c}-\frac{c}{b+c^2}+\frac{1}{a}-\frac{a}{c+a^2}$
$=\sum \frac{1}{a}-\sum \frac{b}{a+b^2}$
Áp dụng BĐT AM-GM:
$\frac{b}{a+b^2}\leq \frac{b}{2\sqrt{ab^2}}=\frac{1}{2\sqrt{a}}$
$\Rightarrow \sum \frac{b}{a+b^2}\leq \sum \frac{1}{2\sqrt{a}}$
$\Rightarrow P\geq \sum \frac{1}{a}-\sum\frac{1}{2\sqrt{a}}$
$=\sum\frac{1}{4}\left(\frac{1}{\sqrt{a}}-1\right)^2+\frac{3}{4}\sum \frac{1}{a}-\frac{3}{4}$
$\geq \frac{3}{4}\sum \frac{1}{a}-\frac{3}{4}$
$\geq \frac{3}{4}.\frac{9}{a+b+c}-\frac{3}{4}$ (theo Cauchy_Schwarz)
$=\frac{3}{4}.3-\frac{3}{4}=\frac{3}{2}$
Vậy $P_{\min}=\frac{3}{2}$. Dấu "=" xảy ra khi $a=b=c=1$
