Muirhead's inequality

Remarks

Proof idea: suppose $a$ majorizes $b$. Then $b$ can be reached from $a$ by finitely many steps, each of which moves two entries closer together while keeping their sum. For one such step it suffices to compare two variables, where the claim reduces to $x^{s} y^{t} + x^{t} y^{s} \ge x^{s'} y^{t'} + x^{t'} y^{s'}$ for $s + t = s' + t'$ and $\abs{s - t} \ge \abs{s' - t'}$, which follows by factoring. Conversely, letting all $x_i$ but some tend to $0$ or $\infty$ shows that the partial sum conditions of majorization are needed.

Example: $a = (2, 0)$ majorizes $b = (1, 1)$, which gives $\frac{1}{2}(x^2 + y^2) \ge x y$. With $a = (n, 0, \dots, 0)$ and $b = (1, 1, \dots, 1)$ one gets $\frac{1}{n}\sum_i x_i^n \ge x_1 x_2 \cdots x_n$, the AM–GM inequality for $x_1^n, \dots, x_n^n$.

Going deeper

For a student

Muirhead's inequality settles many symmetric polynomial inequalities at once, and it is a standard tool in olympiad problems. To use it, write both sides as combinations of symmetric sums $[a]$ and check majorization of the exponent vectors. It does not apply directly when the two sides have different total degree, or when some coefficients are negative; then one combines it with other tools such as Schur's inequality.