A simple Inequality?

Question

For $a,b,c \geq 0$ show that \begin{eqnarray*} 3(a^4+b^4+c^4) +abc(a+b+c) \geq 4(a^2b^2+b^2c^2+c^2a^2). \end{eqnarray*} I have managed to find a proof of this, but it is gruesome & I will post my solution shortly. I am really interested in seeing some more elegant solutions.

I came across this question while trying to solve this question prove this inequality by $abc=1$

Any comments & slick solutions are welcome.

Answer 1

It is clear that \begin{eqnarray*} 3(a^2(2a-b-c)^2+b^2(2b-a-c)^2+c^2(2c-a-b)^2)+ 12(ab(a-b)^2+bc(b-c)^2+ca(a-c)^2)+ a^2(b-c)^2+b^2(c-a)^2+c^2(a-b)^2 \geq 0. \end{eqnarray*} Now rearrange and divide by $4$ and the result follows.

Alternatively cut & paste the following into reduce (3*(a^2*(2*a-b-c)^2+b^2*(2*b-a-c)^2+c^2*(2*c-a-b)^2)+ 12*(ab(a-b)^2+bc(b-c)^2+ca(a-c)^2)+ a^2*(b-c)^2+b^2*(c-a)^2+c^2*(a-b)^2);

Answer 2

BY Schur and AM-GM we obtain: $$\sum_{cyc}(3a^4+a^2bc-4a^2b^2)=$$ $$=\sum_{cyc}(a^4-a^3b-a^3c+a^2bc)+\sum_{cyc}(a^4+b^4+a^3b+ab^3-4a^2b^2)\geq0.$$