Using arithmetic mean>geometric mean Prove that if a,b.c are distinct positive integers that
$$a^4+b^4+c^4>abc(a+b+c)$$
My attempt:
I used the inequality A.M>G.M to get two inequalities
First inequality
$$\frac{a^4+b^4+c^4}{3} > \sqrt[3]{a^4b^4c^4}$$
 or
$$\frac{a^4+b^4+c^4}{3} > abc \sqrt[3]{abc}$$ or new --
first inequality
$$\frac{a^4+b^4+c^4}{3abc} > \sqrt[3]{abc}$$
second inequality:
$$\frac{a+b+c}{3} > \sqrt[3]{abc}$$
I am seeing the numbers in the required equation variables here but am not able to manipulate these to get the inequality I want?? Please direct me on which step should I take after this??

I just want to add another way to solve the problem.
$$\begin{align} & a^4 + b^4 + c^4 = (a^2)^2+(b^2)^2+(c^2)^2 \\& \ge a^2 b^2 + b^2c^2 + c^2a^2 \text{ (From Cauchy-Schwarz)}\\&= (ab)^2 + (bc)^2 + (ca)^2 \\ & \ge abbc + bcca + caab\  (\because a^2 + b^2 + c^2 \ge ab + bc + ac) \\ &= abc(a+b+c), \end{align}$$
