Irish Mathematical Olympiad

$$\begin{align*} & 3(a^3 + b^3 + c^3) - (a + b + c)(a^2 + b^2 + c^2) \\ &= 2(a^3 + b^3 + c^3) + a^2(b + c) + b^2(c + a) + c^2(a + b) \\ &= (a^3 + b^3 - a^2b - ab^2) + (b^3 + c^3 - b^2c - c^2b) + (c^3 + a^3 - c^2a - a^2c) \\ &= (a + b)(a - b)^2 + (b + c)(b - c)^2 + (c + a)(c - a)^2 \\ &\geq (a + b - c)(a - b)^2 + (b + c - a)(b - c)^2 + (c + a - b)(c - a)^2 \\ &\geq a(b - c)^2 + b(c - a)^2 + c(a - b)^2. \end{align*}$$
In other words, the stated result holds and there is equality iff $a = b = c$.

Solution 2:
Let $x = a + b - c$, $y = b + c - a$, $z = c + a - b$, so that
$$a = (z+x)/2, \quad b = (x+y)/2, \quad c = (y+z)/2,$$
whence $a + b + c = x + y + z$,
$$\begin{align*} a^2 + b^2 + c^2 &= \frac{1}{2}(x^2 + y^2 + z^2 + xy + yz + zx) = \frac{1}{2}\left(\sum x^2 + \sum xy\right), \\ (a+b+c)(a^2+b^2+c^2) &= \frac{1}{2}\left(\sum x^3 + 2\sum xy(x+y) + 3xyz\right) \end{align*}$$
and
$$a^3 + b^3 + c^3 = \frac{1}{8}\left(2\sum x^3 + 3\sum xy(x+y)\right).$$
Also,
$$\begin{align*} a(b-c)^2 + b(c-a)^2 + c(a-b)^2 \\ &= \frac{1}{8}\left((z+x)(z-x)^2 + (x+y)(x-y)^2 + (y+z)(y-z)^2\right) \\ &= \frac{1}{8}\left(2(x^3 + y^3 + z^3) - xy(x+y) - yz(y+z) - zx(z+x)\right). \end{align*}$$
Hence
$$\begin{align*} (a+b+c)(a^2+b^2+c^2) + a(b-c)^2 + b(c-a)^2 + c(a-b)^2 \\ &= \frac{1}{2}\left(\sum x^3 + \sum xy(x+y) + \frac{3}{2}xyz\right) \\ &\quad + \frac{1}{8}\left((2x^3+y^3+z^3) - xy(x+y) - yz(y+z) - zx(z+x)\right) \\ &= \frac{3}{4}(x^3+y^3+z^3) + \frac{3}{2}xyz + \frac{7}{8}(xy(x+y) + yz(y+z) + zx(z+x)). \end{align*}$$