Prove $(a^2+b^2+c^2)^3 \geqq 9(a^3+b^3+c^3)$

Suppose $a = \max\{a,b,c\}.$ By the AM-GM inequality we have $$9abc(a^3+b^3+c^3) \leqslant \left(ab+ca+\frac{a^3+b^3+c^3}{3a}\right)^3.$$ Therefore, we need to prove $$a^2+b^2+c^2 \geqslant ab+ca+\frac{a^3+b^3+c^3}{3a},$$ equivalent to $$\frac{(2a-b-c)(a^2+b^2+c^2-ab-bc-ca)}{3a} \geqslant 0.$$ which is true.

Because $(a+b+c)(ab+bc+ca) \geqslant 9abc,$ so we will prove stronger inequality $$(a^2+b^2+c^2)^3 \geqslant (a+b+c)(ab+bc+ca)(a^3+b^3+c^3).$$ or $$(a^2+b^2+c^2-ab-bc-ca)^2\sum (a^2+bc)+ \frac{ab+bc+ca}{2} \sum a^2(b-c)^2 \geqslant 0.$$ Done.

Yes, SOS helps: $$(a^2+b^2+c^2)^3-9(a^3+b^3+c^3)=(a^2+b^2+c^2)^3-9abc(a^3+b^3+c^3)=$$ $$=\frac{1}{2}\sum_{cyc}(2a^6+6a^4b^2+6a^4c^2-18a^4bc+4a^2b^2c^2)=$$ $$=\frac{1}{2}\sum_{cyc}(2a^6-a^4b^2-a^4c^2+7a^4b^2+7a^4c^2-14c^4ab-4a^4bc+4a^2b^2c^2)=$$ $$=\frac{1}{2}\sum_{cyc}(a-b)^2((a+b)^2(a^2+b^2)+7c^4-2abc(a+b+c))=$$ $$=\frac{1}{2}\sum_{cyc}(a-b)^2(7c^4-2abc^2-2ab(a+b)c+(a+b)^2(a^2+b^2))\geq0,$$ where the last inequality is true by AM-GM: $$c^4+\frac{1}{8}(a^2+b^2)(a+b)^2\geq c^4+a^2b^2\geq2abc^2$$ and $$6c^4+\frac{7}{8}(a+b)^2(a^2+b^2)\geq2ab(a+b)c.$$ Can you prove the last inequality by AM-GM?