Đáp án:
$\rm \text{Áp dụng BĐT bunhia:}\\(a^3+b^2+c)\Big(\dfrac{1}{a}+1+c\Big) \geq (a+b+c)^2=9\\\to (a^3+b^2+c) \geq \dfrac{9}{\dfrac{1}{a}+1+c}\\\to \dfrac{a}{a^3+b^2+c} \leq \dfrac{a\Big(\dfrac{1}{a}+1+c\Big)}{9}=\dfrac{a+ac+1}{9}\\CMTT:\dfrac{b}{b^3+c^3+a} \leq \dfrac{ab+b+1}{9}\\\dfrac{c}{c^3+a^2+b} \leq \dfrac{bc+c+1}{9}\\\to \dfrac{a}{a^3+b^2+c}+\dfrac{b}{b^3+c^3+a}+\dfrac{c}{c^3+a^2+b} \leq \dfrac{ab+bc+ca+3+3}{9} \leq \dfrac{3+3+3}{9}=1\\\text{Dấu "=" xảy ra khi:}\\a=b=c=1$
