Từ giả thiết $1={{a}^{2}}+{{b}^{2}}+{{c}^{2}}\ge \dfrac{{{(a+b+c)}^{2}}}{3}$ suy ra $(a+b+c)\le \sqrt{3}$.
Đồng thời $1={{\left[ \sqrt{a}(a\sqrt{a})+\sqrt{b}(b\sqrt{b})+\sqrt{c}(c\sqrt{c}) \right]}^{2}}\le (a+b+c)({{a}^{3}}+{{b}^{3}}+{{c}^{3}})\le \sqrt{3}({{a}^{3}}+{{b}^{3}}+{{c}^{3}})$.
hay $-({{a}^{3}}+{{b}^{3}}+{{c}^{3}})\le -\dfrac{1}{\sqrt{3}}$.
Khi đó $a(2bc-1)=a(2bc-{{b}^{2}}-{{c}^{2}}-{{a}^{2}})=a\left[ -{{a}^{2}}-{{(b-c)}^{2}} \right]\le -{{a}^{3}}$.
Do đó $a(2bc-1)+b(2ca-1)+c(2ab-1)\le -({{a}^{3}}+{{b}^{3}}+{{c}^{3}})\le \dfrac{-1}{\sqrt{3}}$.
Suy ra $M\le 3\sqrt{3}-\dfrac{1}{\sqrt{3}}=\dfrac{8\sqrt{3}}{3}$.
Vậy $\max M=\dfrac{8\sqrt{3}}{3}$ khi $a=b=c=\dfrac{1}{\sqrt{3}}$.