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??
 A: 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.
A: 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
$$\frac12a^4+\frac14b^4+\frac14c^4\ge a^2bc$$
A: an other way $$2a^4+b^4+c^4\geq 4\sqrt[4]{a^8b^4c^4}=4a^2bc$$ $$2b^4+a^4+c^4\geq 4\sqrt[4]{b^8a^4c^4}=4b^2ac$$ $$2c^4+a^4+b^4\geq 4\sqrt[4]{c^8a^4b^4}=4c^2ab$$ adding this ineqalitis we get desired one 
A: Late answer since it was tagged as duplicate there and this way of solving is neither here nor there.
\begin{eqnarray*} abc(a+b+c)
& \stackrel{GM-AM}{\leq} & \left(\frac{a+b+c}{3}\right)^3(a+b+c)\\
& = & 3\left(\frac{a+b+c}{3}\right)^4 \\
& \stackrel{x^4 \; is  \; convex}{\leq} & 3\cdot \frac{1}{3}(a^4+b^4+c^4) = a^4+b^4+c^4
\end{eqnarray*}
Note that the inequality becomes sharp ("<"), if at least two of the numbers are different.
A: I just want to add another way to solve the problem.
$$\begin{align} & a^4 + b^4 + c^4 = (a^2)^2+(b^2)^2+(c^2)^2 \\& \ge a^2 b^2 + b^2c^2 + c^2a^2 \text{ (From Cauchy-Schwarz)}\\&= (ab)^2 + (bc)^2 + (ca)^2 \\ & \ge abbc + bcca + caab\  (\because a^2 + b^2 + c^2 \ge ab + bc + ac) \\ &= abc(a+b+c), \end{align}$$
