Áp dụng bất đẳng thức $\frac{1}{x}+\frac{1}{y}\geq \frac{4}{x+y}$ với $x,y $ dương ta có$\frac{1}{a^2}+\frac{1}{bc}\geq\frac{4}{a^2+bc}\Rightarrow\frac{a}{a^2+bc}\leq\frac{1}{4}(\frac{a}{a^2}+\frac{a}{bc})=\frac{1}{4}(\frac{1}{a}+\frac{a}{bc})$
$\Rightarrow M\leq\frac{1}{4}((\frac{1}{a}+\frac{1}{b}+\frac{1}{c})+(\frac{a}{bc}+\frac{b}{ac}+\frac{c}{ab}))$
$\leq\frac{1}{4}(\frac{ab+bc+ca}{abc}+\frac{a^2+b^2+c^2}{abc})$
$\leq\frac{1}{4}(\frac{a^2+b^2+c^2}{abc}+\frac{a^2+b^2+c^2}{abc})\leq\frac{1}{2}$
Vậy $Max M=\frac{1}{2}$ khi $a=b=c=3$