Reducing an inequality to Schur's inequality Given this,
$a\left(a+b\right)\left(a+c\right)+b\left(a+b\right)\left(b+c\right)+c\left(a+c\right)\left(b+c\right)\ge 2ab\left(a+b\right)+2bc\left(b+c\right)+2ac\left(c+a\right)$
how do you obtain Schur's inequality?
It is suppose to reduce to the nice form of Schur's inequality with $t=1$:
$a(a-b)(a-c)+b(b-a)(b-c)+c(c-a)(c-b)\ge0$
Seems trivial but I can't figure it out.

RHS can be rearranged as $$2ab\left(a+b\right)+2bc\left(b+c\right)+2ac\left(c+a\right)$$
$$=2a^2(b+c)+2b^2(c+a)+2c^2(a+b)$$
Now, observe that $$a(a+b)(a+c)-2a^2(b+c)=a(a-b)(a-c)$$
and you are done!
