When to reach for the Ravi substitution (and when not to)

An inequality in three positive reals is a problem. An inequality in the three sides of a triangle is the same problem carrying three extra constraints that you must use and cannot easily hold on to. That gap is the whole reason the Ravi substitution exists, and it is why knowing the substitution is worth less than knowing the moment to reach for it.

This is not an explanation of what the substitution is. The technique page states it in two lines. This is about the decision: which problems it clears, how to run it without losing marks, and the three situations where reaching for it makes your life worse.

The constraint it deletes

Being the sides of a triangle is a real restriction, not a formality. Take every triple $(a,b,c)$ with each entry a whole number from 1 to 10. There are 1000 of them, and exactly 505 are triangles. Call it a coin flip. Half of the positive triples you could write down are not admissible, and a proof that never notices the difference is proving something false.

The three inequalities $a+b>c$, $b+c>a$, $c+a>b$ carve a cone out of the positive octant. Working inside a cone is awkward: you cannot apply AM-GM to $a$, $b$, $c$ directly and expect the equality case to land anywhere meaningful, because the point where all three are equal is interior, and the points where the inequality is tight sit on the boundary you keep forgetting about.

Ravi reparametrises the cone so that it becomes the whole octant. Write $a = y+z$, $b = z+x$ and $c = x+y$. Then the three triangle conditions become $x>0$, $y>0$ and $z>0$, and that is the entire content of the move: a constrained problem turned into an unconstrained one.

Those three quantities are not arbitrary. If $s$ is the semiperimeter, then $x = s-a$, $y = s-b$ and $z = s-c$, so $x$, $y$, $z$ are the tangent lengths from the vertices to the incircle. Once you have drawn that picture the substitution stops feeling like a trick.

The signals

Three tells, in descending order of reliability.

• The hypothesis is literally "let these be the sides of a triangle". No further thought required. If the problem gives you the triangle condition and then asks for an inequality, it gave you the condition because you need it, and Ravi is the standard way to spend it.
• The expression already carries the tangent lengths. Factors such as $b+c-a$ are $2x$, $2y$, $2z$ on the nose. An expression built out of them is already written in the substituted variables and is asking to be read that way.
• You are stuck because the equality case is unreachable. You have a chain of bounds that is individually correct and collectively too weak, and the reason is that you bounded across the cone boundary. Moving to $x,y,z$ often puts the equality case back at "all three equal", where AM-GM and Cauchy-Schwarz can actually reach it.

The counter-signal is just as useful: if the statement is true for all positive reals, the triangle condition is decoration and substituting only makes the algebra worse. Test a non-triangle triple before you commit. If the claim survives, do not substitute.

Watching it clear a paper

IMO 1964, Problem 2 asks for $a^2(b+c-a) + b^2(c+a-b) + c^2(a+b-c) \le 3abc$ over triangle sides. Signal two fires immediately: all three bracketed factors are the tangent lengths in disguise.

Substitute, and the quantity you actually want to prove non-negative is $$3abc - \big[a^2(b+c-a) + b^2(c+a-b) + c^2(a+b-c)\big].$$ Expand it in $x,y,z$ and everything collapses to $$x^2y + x^2z + y^2x + y^2z + z^2x + z^2y - 6xyz \ge 0.$$ That is AM-GM on six terms whose product is $x^6y^6z^6$, so their arithmetic mean is at least $xyz$. Done, with equality exactly when $x=y=z$, which is the equilateral triangle.

Look at what happened to the difficulty. The original statement is a constrained inequality that resists every standard tool. The substituted one is the first inequality anybody learns, applied to six terms. None of the cleverness is in the finish; all of it was in the change of variables.

When the shape survives, and when it does not

IMO 1983, Problem 6 wants $a^2b(a-b) + b^2c(b-c) + c^2a(c-a) \ge 0$. Substituting turns the left side into $$2\big(x^3z + y^3x + z^3y - xyz(x+y+z)\big),$$ and dividing through by $xyz$ leaves $\frac{x^2}{y} + \frac{y^2}{z} + \frac{z^2}{x} \ge x+y+z$, which Cauchy-Schwarz in Engel form settles in one line.

But notice what did not change. The original is cyclic, not symmetric: at $(a,b,c) = (2,3,4)$ the expression is 16, and swapping the first two sides to $(3,2,4)$ gives 34. Ravi preserves that asymmetry exactly, so the substituted problem is cyclic too, and you still may not assume $x \ge y \ge z$. People lose this problem by substituting, feeling the relief, and then ordering the variables as if the expression were symmetric.

The same problem shows why the triangle condition cannot be skipped. At $(a,b,c) = (1,4,6)$ the expression is $-24$. Those are not triangle sides, and the statement is simply false off the cone. A solution that never invokes the hypothesis is wrong no matter how clean the algebra looks.

What it costs

Three things, and they are the reason this is a decision rather than a reflex.

1. Term count. Expanding $(y+z)^2$ everywhere multiplies the work. A degree-three symmetric expression in $a,b,c$ can become a dozen monomials in $x,y,z$, and expansion errors are the single most common way these problems are lost. Expand slowly and check one numeric value at the end.
2. Geometric meaning. Angles, area and circumradius do not survive into $x,y,z$ in any convenient form. If the problem mixes sides with angles, Ravi strips away the half you still need.
3. Degenerate cases. Ravi as stated gives strictly positive $x,y,z$, so it covers non-degenerate triangles only. If the problem allows a degenerate triangle, you want $x,y,z \ge 0$ and you have to say so.

What to do with this

Run the recognition drill rather than the theory. Take ten inequality problems whose hypothesis mentions a triangle and, for each, write three lines only: does the triangle condition actually matter, are the factors $b+c-a$ present, and what does the equality case look like in $x,y,z$. Do not solve them. The decision is the skill you are short of, and it is trainable in a fraction of the time a full solution takes.

Lemma treats this as a named technique with a place in the sequence rather than a trick you meet twice. The Ravi substitution page sits alongside the classical inequalities it hands off to, the problem archive carries both papers above with difficulty and expected time, and the daily problem posts a fresh one every day in three tiers with no account needed.