Show that $\frac {a+b+c} 3\geq\sqrt[27]{\frac{a^3+b^3+c^3}3}$.
since $$(a+b+c)^3=a^3+b^3+c^3+3(a+b)(b+c)(a+c)=a^3+b^3+c^3+24$$ so $$(a+b+c)^3=a^3+b^3+c^3+3+3+3+3+3+3+3+3\ge 9\sqrt[9]{(a^3+b^3+c^3)\times 3^8}$$
so $$\dfrac{a+b+c}{3}\ge\sqrt[27]{\dfrac{a^3+b^3+c^3}{3}}$$