Schur's Inequality

a^3 + b^3 + c^3 + 3abc ≥ ab(a+b) + bc(b+c) + ca(c+a) — tight at a=b=c and at (a,a,0).

The idea

Theorem (Schur's inequality).

For $a, b, c \ge 0$ and any $t \ge 0$, $a^{t}(a-b)(a-c) + b^{t}(b-a)(b-c) + c^{t}(c-a)(c-b) \;\ge\; 0.$ Equality holds when $a = b = c$, and also when two of the variables are equal and the third is $0$.

At $t = 1$ the inequality expands to the form usually quoted, $a^{3} + b^{3} + c^{3} + 3abc \;\ge\; ab(a+b) + bc(b+c) + ca(c+a).$

The second equality case distinguishes Schur from the other standard symmetric inequalities, which are tight only at the balanced point $a = b = c$. A target inequality that also becomes tight on the boundary, where one variable vanishes, cannot follow from bounds that are strict there. Schur is tight in both places, and the $3abc$ term is responsible.

The hypothesis $a, b, c \ge 0$ cannot be dropped: for variables of mixed sign the inequality fails.

Ways to work on it

• Walkthrough. Statement, hypothesis, and equality cases.
• Practice. Verify Schur on random non-negative triples.
• Hardest. Recognize the inequalities where Schur is the right tool.

Not sure where to start? Take the ten-question placement test.