Prove that $ a^4b^4+ a^4c^4+b^4c^4\le3$

Question

Let $a,b,c>0$ and $a^3+b^3+c^3=3$. Prove that $$ a^4b^4+ a^4c^4+b^4c^4\le3$$

My attempts:

1) $$ a^4b^4+ a^4c^4+b^4c^4\le \frac{(a^4+b^4)^2}{4}+\frac{(c^4+b^4)^2}{4}+\frac{(a^4+c^4)^2}{4}$$

2) $$a^4b^4+ a^4c^4+b^4c^4\le3=a^3+b^3+c^3$$

Answer 1

By AM-GM and Schur we obtain $$\sum_{cyc}a^4b^4=\frac{1}{3}\sum_{cyc}(3ab)a^3b^3\leq\frac{1}{3}\sum_{cyc}(1+a^3+b^3)a^3b^3=$$ $$=\frac{1}{3}\sum_{cyc}(a^3b^3+a^6b^3+a^6c^3)=\frac{1}{9}\sum_{cyc}a^3\sum_{cyc}a^3b^3+\frac{1}{3}\sum_{cyc}(a^6b^3+a^6c^3)=$$ $$=\frac{1}{9}\sum_{cyc}(4a^6b^3+4a^6c^3+a^3b^3c^3)=$$ $$=\frac{1}{9}\left(\sum_{cyc}(a^9+3a^6b^3+3a^6c^3+2a^2b^2c^2)-\sum_{cyc}(a^9-a^6b^3-a^6c^3+a^3b^3c^3)\right)\leq$$ $$\leq\frac{1}{9}\sum_{cyc}(a^9+3a^6b^3+3a^6c^3+2a^2b^2c^2)=\frac{1}{9}(a^3+b^3+c^3)^3=3.$$