Schur's inequality: when to use it, and how to tell AM–GM will not do

Schur's inequality is easy to state and easy to prove. What students get wrong is when to reach for it. They spend twenty minutes trying to finish a symmetric cubic with AM–GM, get within one term of the answer, and never find out that no amount of AM–GM was ever going to close the gap.

So this is about recognition: what a Schur problem looks like, the one test that tells you AM–GM will not be enough, and how to tell when the problem only looks like Schur.

For non-negative reals $a, b, c$ and any $t > 0$,

$$a^t(a-b)(a-c) + b^t(b-a)(b-c) + c^t(c-a)(c-b) \ge 0,$$

with equality when $a = b = c$, and also when two of the variables are equal and the third is $0$.

That second equality case is what gives a Schur problem away.

The proof, because you should be able to write it

Since the expression is symmetric, you may assume $a \ge b \ge c$. Group the first two terms:

$$(a-b)\big[a^t(a-c) - b^t(b-c)\big].$$

Here $a - b \ge 0$, and $a^t \ge b^t$ and $a - c \ge b - c \ge 0$, so the bracket is non-negative too. The third term is $c^t$ times $(c-a)(c-b)$, a product of two non-positive numbers, so it is non-negative as well. That is the whole proof. Notice where non-negativity was used: in $c^t \ge 0$ and in comparing $a^t$ with $b^t$.

The forms it shows up in

You will almost never see Schur written the way it is above. With $t = 1$, three other forms are the same statement, and you should recognise all of them.

• Expanded. $a^3 + b^3 + c^3 + 3abc \ge a^2b + a^2c + b^2a + b^2c + c^2a + c^2b$. Cubes and $abc$ on the big side, the six mixed terms on the small side.
• As a product. $abc \ge (a+b-c)(b+c-a)(c+a-b)$. If a problem compares $abc$ with that product, it is Schur, nothing more.
• In symmetric sums. With $p = a+b+c$, $q = ab+bc+ca$ and $r = abc$, it reads $p^3 - 4pq + 9r \ge 0$. This is the form to use when a problem fixes the sum, or the sum and $q$, and asks about $abc$.

With $t = 2$ you get a degree-4 version, $a^4 + b^4 + c^4 + abc(a+b+c) \ge a^3b + a^3c + b^3a + b^3c + c^3a + c^3b$. In practice $t = 1$ is the one you will use most.

The test: check one boundary point

Here is the test that saves the twenty minutes. Before you try to prove a symmetric inequality, plug in $a = b = 1$, $c = 0$.

If the two sides come out equal, you have a second equality case, and that rules out the usual AM–GM approach. The standard AM–GM bounds for cubics, $a^3 + b^3 + c^3 \ge 3abc$ and the six mixed terms being at least $6abc$, are both slack at $(1, 1, 0)$: each side beats $abc = 0$ by $2$. A chain of steps that are strictly loose at $(1, 1, 0)$ cannot prove an inequality that is exactly tight there. That is the moment to switch to Schur.

Compare the inequality $a^2 + b^2 + c^2 \ge ab + bc + ca$. At $(1, 1, 0)$ the left side is $2$ and the right side is $1$, so it is not tight. Its only equality case is $a = b = c$, and it is half the sum of $(a-b)^2$, $(b-c)^2$ and $(c-a)^2$. One equality case usually means the classical inequalities are enough. A second equality case at a boundary point is what calls for Schur.

The second thing to check is the hypothesis. Schur needs non-negative variables. At $a = b = 0$, $c = -1$ the $t = 1$ expression is $-1$. That is also why no sum-of-squares argument can prove it. A sum of squares is non-negative for every real input, and this expression is not.

Watching it work: IMO 1964, Problem 2

IMO 1964, Problem 2: if $a, b, c$ are the sides of a triangle, prove

$$a^2(b+c-a) + b^2(c+a-b) + c^2(a+b-c) \le 3abc.$$

Run the test. At $(1, 1, 0)$ the left side is $0$ and so is $3abc$: tight. That points to Schur, and it is Schur. Expand the left side: it is the six mixed terms minus $a^3 + b^3 + c^3$. Move it across and the claim becomes exactly the expanded form above. The proof is one line.

Notice what the triangle condition did there: nothing. The inequality holds for every triple of non-negative reals, triangle or not. At $(5, 1, 1)$, which is no triangle, the left side is $-65$ and the right side is $15$. The hypothesis is there because the problem was written about triangles, and the only equality case that is an actual triangle is the equilateral one.

There is a second route, and it is worth seeing because it shows when Schur is not needed. The Ravi substitution $a = y+z$, $b = z+x$, $c = x+y$ with $x, y, z > 0$ uses the triangle condition properly. Then $b + c - a = 2x$ and so on, and after expanding, $3abc$ minus the left side is the six mixed terms in $x, y, z$ minus $6xyz$. That is plain AM–GM, with no Schur at all. In $x, y, z$ the tight point $(1, 1, 0)$ has moved out of reach: a degenerate triangle needs one of $x, y, z$ to be $0$, and the substitution keeps all three positive.

So the same problem can be pure Schur or pure AM–GM, depending on the variables you write it in. The test tells you which: check $(1, 1, 0)$ in whatever variables you are about to work with.

Using the symmetric-sum form

When a problem fixes some of the symmetric sums, the symmetric-sum form turns Schur into a direct bound. Say $a + b + c = 6$ and $ab + bc + ca = 11$, with all three non-negative. Then $p^3 - 4pq = 216 - 264 = -48$, so Schur says $9abc \ge 48$, that is, $abc \ge \tfrac{16}{3}$.

That bound is valid but not sharp: the true minimum is about $5.62$, reached when two of the variables are equal. Schur's own equality cases, all equal or one of them zero, are not possible with this $p$ and $q$, so the bound cannot be reached. Remember this when a problem asks for the exact minimum. Schur gives a lower bound quickly, and you still have to check whether it is achieved.

What to do with this

Make the $(1, 1, 0)$ test automatic. Every time you meet a symmetric inequality in three variables, write the value at $a = b = c$ and the value at $(1, 1, 0)$ before anything else. One equality case: try AM–GM or Cauchy–Schwarz first. Two: write Schur in whichever of the four forms matches, and see if what is left is a single obvious step.

The problem archive has IMO 1964/2 with hints and a full solution, and the techniques pages show where AM–GM, Cauchy–Schwarz and Ravi sit in the ladder. The daily problem is a free way to keep practising between sessions.