Schur's inequality

Remarks

Proof idea: the expression is symmetric, so assume $x \ge y \ge z$. Group the first two terms as $(x - y)\big(x^t (x - z) - y^t (y - z)\big)$, which is $\ge 0$ because $x^t \ge y^t \ge 0$ and $x - z \ge y - z \ge 0$. The last term $z^t (x - z)(y - z)$ is also $\ge 0$.

Example: for $t = 1$, expanding gives $$x^3 + y^3 + z^3 + 3xyz \ge xy(x + y) + yz(y + z) + zx(z + x).$$ In terms of the elementary symmetric polynomials $p = x + y + z$, $q = xy + yz + zx$ and $r = xyz$, this reads $p^3 - 4pq + 9r \ge 0$.

Going deeper

For a student

Schur's inequality is a degree $t + 2$ symmetric inequality that does not follow from Muirhead's inequality alone, because it has negative coefficients. It is a standard companion of Muirhead's inequality in proofs of symmetric inequalities in three variables. Its equality cases, which include $(1, 1, 0)$, show that it is sharp at the boundary as well as at the centre.