When a smoothing argument works, and when it quietly lies

Smoothing is the shortest correct proof of a symmetric inequality, and it is also the most common source of a proof that is wrong in a way nobody notices. Both facts come from the same place: the argument is three sentences long, and two of them are usually missing.

This is not an explanation of what smoothing is. It is the checklist that decides whether the smoothing argument you are about to write is a proof or a guess with algebra attached.

Smoothing is a move, not a theorem

You do not cite smoothing. You perform it. Pick two of the variables, move them toward each other while holding the constraint fixed, and ask what happened to the expression. If it improved, the extremum cannot be at a point where those two differ, so the extremum sits where everything is equal.

Written out, the argument has exactly four obligations, and a write-up that skips any of them is incomplete:

1. The move. Name it in formulas. Not "push $a$ and $b$ together": replace $(a, b)$ by $(a + t,\ b - t)$ for $0 < t \le \tfrac{b-a}{2}$.
2. The constraint survives. Show the move stays on the constraint surface. This is one line and it is the line that decides which move you are allowed.
3. The expression improves. Strictly, or you learn nothing about where the extremum is, only where it might be.
4. The process terminates. Either finitely many moves reach the equal point, or you argue that a maximum exists at all and that it cannot be at an unequal point.

Obligation 4 is the one that separates a competition proof from a plausible paragraph. Nudging toward equality does not, by itself, prove that the equal point is where the extremum lives. It proves that no unequal point is the extremum, which is only the same statement once you know an extremum exists.

The two engines

Almost every smoothing step in practice runs on one of two elementary facts, and it is worth knowing which one your constraint hands you.

A fixed sum. If $a + b$ is held constant and the two are moved toward each other, the product strictly increases and the sum of squares strictly decreases. That is the whole engine behind "for fixed perimeter, the equilateral case wins". Any expression built out of a constant sum and a product is going to move in a direction you can predict.

A fixed product. If $ab$ is held constant and the two are moved toward each other, their sum strictly decreases, bottoming out at $2\sqrt{ab}$ when they meet. Constraints of the form $abc = 1$ give you this move, and the equality case they force is exactly the point where all the variables collapse onto one another.

Both are one line of algebra and both are worth re-deriving once by hand, because the direction is the entire content. If you cannot say in advance whether your move raises or lowers the expression, you are not ready to write the proof.

Normalise before you smooth

Most inequality problems are homogeneous, meaning every term has the same degree, and scaling all the variables by $t$ scales the whole expression by a fixed power of $t$. When that is true you may simply declare $a + b + c = 3$ or $abc = 1$, which removes a degree of freedom for nothing. Normalisation is almost always the right first line, and it frequently converts a problem with no obvious smoothing move into one with an obvious one.

The obligation attached is a real one, not a formality. Check the degree. The expression in the IMO 2001 inequality is homogeneous of degree $0$, so every scaling leaves it untouched and any normalisation at all is free. Something like $a + b + c + abc$ is not homogeneous of any degree: doubling all three variables multiplies it by $3.5$ at one point and by $2$ at another, so there is no exponent that describes what scaling does. Declaring $a + b + c = 3$ there is a genuine loss of generality, and the proof that follows is about a different problem.

There is a second kind of normalisation worth knowing, which trades a hard constraint for an easy one rather than removing it. When the variables are the sides of a triangle, the Ravi substitution rewrites them as $a = y + z$, $b = z + x$, $c = x + y$, and the triangle condition becomes nothing more than $x, y, z > 0$. The IMO 1964 problem is the standard demonstration: unusable as stated, routine once the triangle condition is spent.

The first way it lies: the extremum is on the boundary

Take $x + y + z = 3$ with all three non-negative, and look at $x^2 + y^2 + z^2$. Moving two variables toward each other lowers it, every time, so the minimum is at $x = y = z = 1$, where the value is $3$. That is a correct smoothing argument.

Now run the same words for the maximum. There is no contradiction to find, because the maximum is not at an interior point at all: it is $9$, attained when two of the variables are pushed all the way to zero. Smoothing moved you toward equality, and the maximum was in exactly the opposite direction, sitting on the boundary of the region.

Boundary extrema are why a smoothing proof must state the constraint region, including whether the variables may reach zero. An inequality over strictly positive reals and the same inequality over non-negative reals can have their extrema in completely different places.

The second way it lies: $n - 1$ equal, not $n$

The more expensive failure is subtler, because the argument survives the direction test and still concludes something false.

Consider the symmetric expression $a(a-b)(a-c) + b(b-a)(b-c) + c(c-a)(c-b)$ for non-negative $a, b, c$. It is never negative, so zero is its minimum, and a smoothing argument will happily tell you the minimum is at $a = b = c$. It is. It is also at $(1, 1, 0)$, where the expression is zero again. Any write-up concluding "equality holds only when all three are equal" has stated something false, and on a paper that asks for the equality case, that is marks gone.

The general shape behind this: for many constrained symmetric problems the extremum has $n - 1$ variables equal and one different, not all $n$ equal. A smoothing step that moves two variables together proves nothing about the odd one out. So when the problem asks for the equality case, hunt for it before you write, by testing the points where one variable goes to the boundary while the rest stay equal.

A recognition pass on a real problem

The IMO 1995 inequality asks for a lower bound of $\tfrac32$ under the constraint $abc = 1$. Run the checklist cold:

• Symmetric under swapping any two variables, so a two-variable move is available.
• The constraint is a fixed product, so the legal move is the fixed-product one: replace $(a, b)$ by $(\sqrt{ab}, \sqrt{ab})$, which keeps $abc$ where it was.
• The equal point is $a = b = c = 1$, and the expression there is exactly $\tfrac32$, which matches the bound asked for. That agreement is the signal that the extremum really is at equality and that a smoothing argument has somewhere to land.
• Variables are strictly positive, and pushing any of them toward zero sends the expression up, not down. So the minimum is interior and the boundary trap does not apply.

Four bullets and you know the argument is viable before writing a line of it. Note what the recognition pass did not do: it did not prove the inequality. On this particular problem the shortest complete proof is not a smoothing argument at all, but the pass told you what the answer must be, and knowing the target is most of the work.

What to practise

The skill being trained here is not algebra, it is the habit of writing the four obligations down before trusting the conclusion. Take ten symmetric inequality problems and, for each, write only: the move, why the constraint survives, the direction, and where the equality case sits. Do not finish them. You will learn more per minute than by solving three.

Lemma treats smoothing and normalisation as one named technique rather than a trick that shows up twice and is never explained. The problem archive carries the competition inequalities above in full, each with a marking rubric so a proof can be scored honestly rather than declared finished. And the daily problem posts a fresh one every day at three difficulties, no account needed.