The following quantity is clearly positive
\begin{eqnarray*}
3a^6(a^3-b^3)^2+
4a^4(a^4-b^2c^2)^2+
3b^6(a^3-b^3)^2+
6b^4(b^4-a^2c^2)^2+
3a^2c^2(b^4-a^2c^2)^2+
2^6c^2(ab-c^2)^2+
4c^6(a^3-b^3)^2+
3b^6c^2(ab-c^2)^2+
4a^3b(b^4-a^2c^2)^2+
6a^4b^4(c^2-ab)^2+
8a^3b^3c^2(c^2-ab)^2.
\end{eqnarray*}
This can be rearranged to
\begin{eqnarray*}
7a^{12}-6a^9b^3+12a^6b^6-2a^3b^9+9b^{12} -6a^8b^2c^2-12a^5b^5c^2-6a^2b^8c^2-6a^4b^4c^4-6ab^6c^5+9a^6c^6+7b^6c^6.
\end{eqnarray*}
Now setting this to be greater than or equal to zero & rearranging some more and we have
\begin{eqnarray*}
2(a^4+ab^3+b^2c^2)^3 \leq 9(a^6+b^6)(a^6+b^6+c^6).
\end{eqnarray*}
Multiply by $a^6$, divide by $54$ and take the cube root
\begin{eqnarray*}
\frac{ a^6+a^3b^3+a^2b^2c^2 }{3} \leq \sqrt[3]{ a^6 \left( \frac{a^6+b^6}{2} \right) \left(\frac{a^6+b^6+c^6}{3} \right)}.
\end{eqnarray*}
The result then follows by replacing $a^6 \rightarrow a$ , $b^6 \rightarrow b$ & $c^6 \rightarrow c$.