<s>[INST] 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??
 [/INST] Here is a slightly more general approach to construct the right AM-GM.  Note, for non-negative $x,y$ s.t. $x+y\le 1$,
$$x a^4+yb^4+(1-x-y)c^4\ge a^{4x}b^{4y}c^{4-4x-4y}$$
Clearly we are done if we have $4x=2,4y=1\implies 4-4x-4y=1$, so $x=\frac12, y=\frac14$ and what you need is to sum
$$\frac12a^4+\frac14b^4+\frac14c^4\ge a^2bc$$
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