When you can write WLOG, and when it breaks the proof

"WLOG" is three letters that save half a page, and it is one of the most reliable places an otherwise correct solution loses marks. Rarely because the assumption was wild. Usually because the line was written out of habit: the expression looked symmetric, the ordering got assumed, and nobody stopped to ask whether the problem permitted it.

The fix is not to write WLOG less often. It is to know, in one sentence, what licenses it, and to be able to say that sentence out loud before the ink dries.

The test

A "without loss of generality" is legal exactly when you can name a map on the configurations with two properties:

1. It reaches your assumption. Every configuration is sent by the map to one that satisfies the thing you are about to assume.
2. It preserves the truth. The statement holds for a configuration if and only if it holds for its image.

Property 1 without property 2 is a lie, and property 2 without property 1 is useless. Both together mean the cases you dropped are copies of the cases you kept, which is the entire content of the phrase.

The map is usually so obvious that naming it feels pedantic. Name it anyway. The one time it is not obvious is the time it does not exist, and that is precisely the time you will not notice.

Three licences you will actually use

Full symmetry gives you an ordering. If an expression is unchanged under all six permutations of $a$, $b$, $c$, the map is "sort the triple", and you may assume $a \ge b \ge c$. Schur's inequality is the clean case: $x(x-y)(x-z) + y(y-x)(y-z) + z(z-x)(z-y) \ge 0$ takes the same value under every one of the six relabellings, so the ordering is free, and once you have it the proof is two lines of grouping. Schur is what IMO 1964, Problem 2 collapses to after the Ravi substitution, which is worth noticing on its own: the substitution was chosen partly because it restores the symmetry the triangle condition was hiding.

Relabelling gives you a name, including a colour. In the six-person party problem the standard proof fixes a vertex, finds three edges of one colour, and says "say red". That "say" is a WLOG, and its licence is the map that swaps the two colours: the conclusion asks for a monochromatic triangle in either colour, so swapping sends the problem to itself. Read the six-person proof with that in mind and you will see the word doing real work rather than decorating.

Symmetry of the equation, not of the expression. In IMO 1988, Problem 6, the condition rearranges to $a^2 + b^2 = k(ab+1)$, which is unchanged when $a$ and $b$ trade places. So the set of valid pairs is closed under the swap, and assuming $a \ge b$ inside a minimal counterexample costs nothing. Here the licence lives in the constraint, not in the thing being proved, and that is the flavour most students miss.

The cyclic trap

This is the one that produces confident, well written, wrong proofs.

An expression can be invariant under rotating $a \to b \to c \to a$ without being invariant under swapping two variables. Cyclic is not symmetric, and only the second one licenses an ordering.

Take the claim $a^2b + b^2c + c^2a \ge ab^2 + bc^2 + ca^2$ for positive reals. It looks like an inequality you have seen. The difference factors as $-(a-b)(b-c)(c-a)$, so if you assume $a \ge b \ge c$ the three brackets have signs $+$, $+$, $-$, the product is at most zero, and the negation makes it non-negative. Proof complete, in one line.

The claim is false. Put $a = 1$, $b = 2$, $c = 3$: the left side is $2 + 12 + 9 = 23$ and the right side is $4 + 18 + 3 = 25$. Nothing in the line of algebra was wrong. The WLOG was wrong, and it imported a hypothesis that made a false statement true.

Rotation does buy you something, just less than you wanted. You can always rotate until the largest variable sits first, so $a = \max(a,b,c)$ is free. What you cannot do is choose between $a \ge b \ge c$ and $a \ge c \ge b$. Those two are separate cases, and in a genuine cyclic problem they behave differently.

IMO 1983, Problem 6 is exactly this shape. Its left side $a^2b(a-b) + b^2c(b-c) + c^2a(c-a)$ evaluates to $34$ at $(3,4,5)$ and stays $34$ under rotation, but jumps to $58$ at $(4,3,5)$. Any solution that opens with "the expression is symmetric, so assume $a \ge b \ge c$" has already lost, whatever follows.

Symmetry as a search tool, not just a licence

The other half of the topic is that symmetry tells you where the answer probably lives before you have proved anything. In a symmetric problem the extreme usually sits at $a = b = c$, which is why guessing the equality case first and building the proof toward it works so often.

Treacherously, "usually" is not "always", and the other candidate is the boundary. Minimise $a^2 + b^2$ subject to $a + b = 1$ with $a, b \ge 0$ and you get $\tfrac12$ at the symmetric point $a = b = \tfrac12$. Maximise the same expression under the same constraint and you get $1$, at $a = 1$, $b = 0$, as far from symmetric as the region allows. So "WLOG $a = b$" is never a licence. It is a guess about where to look, and the proof still has to cover everything else.

What to write on the page

Write the symmetry, not the abbreviation. A grader reading "WLOG $a \ge b \ge c$" has to reconstruct your reason; a grader reading "the expression is symmetric in $a$, $b$, $c$, so we may assume $a \ge b \ge c$" has nothing left to check. When the problem is only cyclic, say so and pay the price honestly: "we may assume $a = \max(a,b,c)$; the two orderings of $b$ and $c$ are treated separately below". Then treat them separately below.

Lemma keeps these as two named techniques, because they fail differently: symmetry is how you find the answer, and symmetry breaking and WLOG is how you are allowed to write it down. The problem archive has the papers cited above with hints and full solutions, the daily problem posts three tiers every day with no account needed, and the other guides cover the neighbouring recognition questions.