Prove that $a^2(b+c-a)+b^2(a+c-b)+c^2(a+b-c) \leq 3abc.$
Solution 1:
Since $a, b, c$ are the lengths of the sides of a triangle, we can let $a=x+y, b=y+z, c=z+x$, where $x, y, z>0$. This is known as Ravi Substitution.
Substituting, we have:
$2(x+y)^2z+2(y+z)^2x+2(z+x)^2y\leq3(x+y)(y+z)(z+x)$
Expanding, we have:
$$2\sum_{sym}x^2y+12xyz\leq6xyz+3\sum_{sym}x^2y$$
or
$$6xyz\leq\sum_{sym}x^2y$$
which is clearly true by AM-GM inequality.
Solution 2:
It simplifies to $$a(a-b)(a-c)+b(b-a)(b-c)+c(c-a)(c-b) \ge 0$$ which is Schur's Inequality for $r=1$.