For every three real numbers $a,b,c$ prove that $$a^4+b^4+c^4+a^3(b+c)+b^3(a+c)+c^3(a+c)-3a^2bc-3ab^2c-3abc^2 \geq 0.$$


I was thinking of using AM-GM but that seems like it would make it worse. I rewrote it as $$a^4+b^4+c^4+a^3(b+c)+b^3(a+c)+c^3(a+c)-3a^2bc-3ab^2c-3abc^2 = a^4+b^4+c^4+(b+c)(a^3+(a+c)(b^2-bc+c^2)+(b^3+c^3)(a+c)-3a^2bc-3ab^2c-3abc^2$$ Then I don't see how this helps to show it is greater than or equal to 0.

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    $\begingroup$ Shouldn't that $c^3(a+c)$ be $c^3(a+b)$ given the pattern of others? $\endgroup$ – user64066 Jan 2 '16 at 20:23
  • $\begingroup$ According to my book no, but it could be a typo. $\endgroup$ – Jacob Willis Jan 2 '16 at 20:43
  • $\begingroup$ @JacobWillis: It is a typo, or else it would be false for $a=20$, $b=20$, $c=19$. $\endgroup$ – darij grinberg Jan 2 '16 at 21:16

Assuming the correction suggested by user64066, the left hand side factors as $\left(a+b+c\right)^2\left(a^2+b^2+c^2-bc-ca-ab\right)$. While $\left(a+b+c\right)^2$ is clearly nonnegative, the nonnegativity of $a^2+b^2+c^2-bc-ca-ab$ is well-known (for example, it follows from $2\left(a^2+b^2+c^2-bc-ca-ab\right)=\left(b-c\right)^2+\left(c-a\right)^2+\left(a-b\right)^2 \geq 0$). So the left hand side is nonnegative, qed.


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