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??

You have to apply it repeatedly on different levels.
First, $(a^4+b^4)/2\ge a^2b^2$. Also, $(a^4+c^4)/2\ge a^2c^2$. Now apply the AM-GM to these two.
$${(a^4+b^4)/2+(a^4+c^4)/2\over2}\ge\sqrt{a^2b^2\cdot a^2c^2}=a^2bc=abc\cdot a$$
Now write two other similar inequalities with $a,b,c$ in different order, and add them all together.
