Prove this inequality $a^4+b^4+c^4+9\ge 4(a^2b+b^2c+c^2a)$

Question

let $a,b,c>0$ and such $abc=1$, show that

$$a^4+b^4+c^4+9\ge 4(a^2b+b^2c+c^2a)$$

Schur inequality $$a^4+b^4+c^4\ge \sum_{cyc}ab(a^2+b^2)-abc(a+b+c)$$ It suffices to show that $$\sum_{cyc}ab(a^2+b^2)+9\ge 4(a^2b+b^2c+c^2a)+abc(a+b+c)$$