How to prove an olympiad inequality, step by step

Nobody gets stuck on an olympiad inequality because they have forgotten AM–GM. They get stuck because they picked AM–GM in the first minute, before they knew anything about the inequality, and then spent forty minutes trying to make a tool fit a problem it was never shaped for.

The fix is not a cleverer tool. It is an order of attack in which the tool is chosen fourth, after three cheap steps that tell you which tools are even allowed. Each step takes a minute or two. Skipping them is what costs the forty.

Step 1: find every equality case first

Before anything else, plug in numbers. Start with all variables equal. Then try a boundary: two equal and one at $0$, or one variable large and the others small. Write down every point where the two sides are equal.

This is not a warm-up. The equality case is the specification of the proof. Every inequality you apply along the way has to be an equality at that point too, or the chain finishes strictly loose and proves something weaker than what was asked.

Two examples show what this buys you.

• Nesbitt's inequality, $\frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b} \ge \frac32$, is tight at $a = b = c$. At $(1, 1, 0)$ the left side is $2$, comfortably above. One equality case, in the middle: the classical inequalities are candidates.
• IMO 1964, Problem 2, $a^2(b+c-a) + b^2(c+a-b) + c^2(a+b-c) \le 3abc$, is tight at $a = b = c$ and at $(1, 1, 0)$, where both sides are $0$. A second equality case at the boundary rules out any chain of steps that is loose there, which is most AM–GM chains. You learn that in thirty seconds, before trying any of them.

Step 2: check the degree, then normalise

Count the degree of each term. If every term has the same degree, the inequality is homogeneous: scaling all variables by the same factor changes nothing, so you may choose the scale. Set $a + b + c = 1$, or $3$, or $abc = 1$, whichever makes the next step cleanest.

If the inequality is not homogeneous, there is almost always a constraint, and the constraint is there to make it homogeneous. $abc = 1$ alongside terms of different degrees is a hint to multiply through by powers of $abc$ until the degrees match, or to substitute.

Step 3: substitute until the shape is familiar

Some inequalities are standard ones in disguise. The disguise is usually a reciprocal or a triangle.

IMO 1995, Problem 2 asks you to show $\sum \frac{1}{a^3(b+c)} \ge \frac32$ when $abc = 1$. Put $x = \frac1a$, $y = \frac1b$, $z = \frac1c$, so $xyz = 1$. Then

$$\frac{1}{a^3(b+c)} = \frac{x^3}{\frac1y + \frac1z} = \frac{x^3 yz}{y+z} = \frac{x^2}{y+z},$$

using $xyz = 1$ in the last step. The problem becomes $\sum \frac{x^2}{y+z} \ge \frac32$, which is a shape you have seen before. For triangle sides, the Ravi substitution $a = y+z$, $b = z+x$, $c = x+y$ does the same job.

Step 4: now choose the tool, by shape

Only now. The shape after steps 2 and 3 usually decides it.

• Squares over sums, like $\sum \frac{x^2}{y+z}$: Cauchy–Schwarz in its fraction form, $\sum \frac{x^2}{y+z} \ge \frac{(x+y+z)^2}{2(x+y+z)}$. The AM–GM and Cauchy–Schwarz page covers both.
• A product fixed, a sum wanted (or the reverse): AM–GM. With $xyz = 1$, $x + y + z \ge 3$.
• A sum of one function applied to each variable, with the variables summing to a constant: Jensen's inequality, if the function is convex (or concave, for the other direction). With coefficients in front, use the weighted form.
• None of the above, but the equality case is symmetric: smoothing. Show that moving two variables towards each other does not increase the expression, and the problem collapses to the equal case.

Finish IMO 1995 this way: Cauchy–Schwarz gives $\sum \frac{x^2}{y+z} \ge \frac{x+y+z}{2}$, and AM–GM with $xyz = 1$ gives $\frac{x+y+z}{2} \ge \frac32$. Both steps are equalities at $x = y = z = 1$, the point step 1 found.

Step 5: audit every line twice

For each step, ask two questions.

1. Does it go the right way? You are proving a lower bound, so every replacement must make the expression smaller or keep it equal. The classic slip: in $\frac{x^2}{y+z}$, replacing the denominator by $2\sqrt{yz}$ looks like AM–GM, but $y + z \ge 2\sqrt{yz}$ makes the fraction larger, so $\frac{x^2}{y+z} \le \frac{x^2}{2\sqrt{yz}}$. That is the wrong direction for a lower bound, and it is easy to miss when the step feels familiar.
2. Is it tight at every equality case? Plug the equality case from step 1 into the step. If the step is strict there, the chain cannot reach the target. Stop and change tools rather than pushing on.

Checking one random point numerically after each line takes seconds and catches both mistakes.

Watching it run: IMO 2001, Problem 2

The problem: for positive reals $a, b, c$,

$$\frac{a}{\sqrt{a^2+8bc}} + \frac{b}{\sqrt{b^2+8ca}} + \frac{c}{\sqrt{c^2+8ab}} \ge 1.$$

Step 1. At $a = b = c$ each term is $\frac{a}{\sqrt{9a^2}} = \frac13$, so the sum is $1$: equality. Now the boundary. Take $a = 1$ and let $b$ and $c$ shrink towards $0$: the first term approaches $1$ and the other two approach $0$. So the left side gets as close to $1$ as you like there too. Two places to stay tight, the middle and the corner.

Step 2. Each term has degree $0$, so the inequality is homogeneous. Normalise: $a + b + c = 1$. That choice is not arbitrary. With the sum equal to $1$, the numerators $a, b, c$ are now weights.

Step 3. Nothing to substitute. The shape is already plain.

Step 4. A weighted sum of one function, $f(t) = \frac{1}{\sqrt t}$, evaluated at $t = a^2 + 8bc$ and its cousins, with weights summing to $1$. That is weighted Jensen, and $f$ is convex for $t > 0$. So

$$\sum a \cdot f(a^2 + 8bc) \ge f\Big(\sum a(a^2 + 8bc)\Big) = \frac{1}{\sqrt{a^3 + b^3 + c^3 + 24abc}}.$$

It remains to show $a^3 + b^3 + c^3 + 24abc \le 1 = (a+b+c)^3$. Expanding,

$$(a+b+c)^3 = a^3 + b^3 + c^3 + 3(a^2b + a^2c + b^2a + b^2c + c^2a + c^2b) + 6abc,$$

so what is left is $a^2b + a^2c + b^2a + b^2c + c^2a + c^2b \ge 6abc$. That is AM–GM on six terms whose product is $a^6b^6c^6$.

Step 5. Audit against both equality cases. At $a = b = c$, Jensen is an equality because every argument is the same, and the six-term AM–GM is an equality because every term is the same. At the corner $(1, 0, 0)$, Jensen puts all the weight on one point, so it is an equality, and the last inequality reads $0 \ge 0$. Every step survived both checks, which is how you know the chain was the right one before you write it out.

What to do with this

The order is trained, not memorised. Take ten inequalities you have already solved, and for each one write only the output of steps 1 to 3: the equality cases, the normalisation, the substitution. Do not re-solve them. You will find that in most of them the tool you used was already forced by the time step 3 was done.

The technique pages for AM–GM and Cauchy–Schwarz, Jensen and smoothing have each tool on its own, with worked problems. The problem archive holds all four inequalities above in full, and more guides take single tools one at a time.