AM–GM or Cauchy–Schwarz: how to know which one to use

Two inequalities cover most of what a contest will ask you to prove, and both fit on a single line. That is exactly the difficulty. Nothing in a fresh problem says AM–GM or Cauchy–Schwarz, and choosing wrong costs you ten minutes of perfectly correct algebra that ends in a bound too weak to close.

The choice is not about which inequality is stronger. It is about which conversion the problem needs, and the conversion is visible in the shape of the expression before you write a single line.

What each one actually converts

• AM–GM moves between a sum and a product. $\frac{a_1 + \dots + a_n}{n} \ge \sqrt[n]{a_1 \cdots a_n}$. Read left to right it bounds a sum below by a product; read right to left it bounds a product above by a sum.
• Cauchy–Schwarz moves between a sum of products and a product of sums. $\left(\sum a_i b_i\right)^2 \le \left(\sum a_i^2\right)\left(\sum b_i^2\right)$. Its Engel form, $\sum \frac{a_i^2}{b_i} \ge \frac{\left(\sum a_i\right)^2}{\sum b_i}$, moves between a sum of fractions and one single fraction.

So AM–GM is the tool that knows what a product is, and Cauchy–Schwarz is the tool that knows what a denominator is. Nearly every recognition decision in this subject follows from that one sentence.

The shape test

Three questions, asked in order. Take the first one that fires.

1. Is a product doing real work — in the constraint you are given, or in the quantity you are asked to bound? Reach for AM–GM.
2. Are the denominators sums? Reach for Cauchy–Schwarz in Engel form.
3. Does the left-hand side split into two sequences multiplied term by term? Reach for Cauchy–Schwarz in its plain form.

Signal one: a product is doing the work

Given positive reals with $abc = 1$, prove $a + b + c \ge 3$ is the purest case there is. The constraint is a product and the target is a sum, so the whole problem is one conversion: $\frac{a+b+c}{3} \ge \sqrt[3]{abc} = 1$.

Notice what Cauchy–Schwarz would have to do here. Its statement contains squares and cross terms, and a constraint on $abc$ has nothing to attach to. You can force a route through it, but the constraint has to be spent by AM–GM eventually regardless, so you may as well spend it first.

The same signal fires when the product is in the target rather than the constraint. Prove $(a+b)(b+c)(c+a) \ge 8abc$ is three separate two-term AM–GMs, one per bracket, multiplied together because all three sides are positive. The tell is that a product of sums is being compared to a product of the variables themselves — nothing but a sum-to-product conversion can bridge that.

The signal survives translation into other subjects, which is where it earns its keep. Among all triangles of a given perimeter, prove the equilateral one has the largest area looks like geometry and is not. Heron's formula turns the area into $\sqrt{s(s-a)(s-b)(s-c)}$, and a fixed perimeter means the three factors $s-a$, $s-b$, $s-c$ have a fixed sum. Maximise a product given a fixed sum: that is AM–GM read right to left, and the maximum sits where the factors are equal, which is the equilateral triangle. Question one fired the moment a product appeared with its sum pinned down.

Signal two: the denominators are sums

The moment a sum appears underneath a fraction bar, AM–GM starts working against you. Bounding $b + c \ge 2\sqrt{bc}$ makes the denominator smaller, which makes $\frac{a}{b+c}$ larger, and an upper bound is not what a problem asking for $\ge$ wants. The direction is the whole issue, and it is the single most common wasted page in an inequality attempt.

Engel form exists for precisely this shape, and it takes one preparatory move that is worth learning as a reflex: manufacture the square on top. A term $\frac{a}{b+c}$ is not yet in Engel shape, but $\frac{a^2}{a(b+c)}$ is the same number and is. Do that to all three terms of Nesbitt's inequality and the sum collapses in one step:

$$\frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b} \ =\ \sum \frac{a^2}{a(b+c)} \ \ge\ \frac{(a+b+c)^2}{2(ab+bc+ca)}$$

The denominator is $2(ab+bc+ca)$ because $a(b+c) + b(c+a) + c(a+b)$ counts each product twice. What is left is no longer a sum of fractions at all, which is the entire point of the conversion.

Multiplying by $a$ on top and bottom looks like it accomplishes nothing, and that is why people do not think of it. It is the standard way into Engel form and it costs a line.

The failure worth studying

That same Nesbitt sum is where a student who has only learned AM–GM discovers the limit of it, and the way it fails is more instructive than the way it succeeds. The target is $\frac32$.

Apply AM–GM to the three fractions directly and you get a genuinely valid bound:

$$\frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b} \ \ge\ 3\sqrt[3]{\frac{abc}{(a+b)(b+c)(c+a)}}$$

Now look at what is under that cube root. By the bound above, $(a+b)(b+c)(c+a) \ge 8abc$, so the fraction inside is at most $\frac18$ and the whole right-hand side is at most $3 \cdot \frac12 = \frac32$ — with equality only when $a = b = c$. The AM–GM route produces a bound that reaches $\frac32$ exactly at the point where the problem is already tight, and sits strictly below it everywhere else. It can never prove the statement.

Nothing was wrong with the algebra. The tool simply converted in a direction that threw away the information the problem needed.

The tie-breaker: read the equality case first

When the shape test leaves you genuinely undecided, the equality case decides, and it decides quickly because the two tools have different equality conditions.

• AM–GM is tight exactly when all the averaged terms are equal.
• Cauchy–Schwarz is tight exactly when the two sequences are proportional — that is, when one is a scalar multiple of the other.

So: find where the problem's own equality lives before you bound anything. For a symmetric inequality that is usually $a = b = c$, but it is worth checking rather than assuming, because an asymmetric constraint moves it. Then choose the grouping whose equality condition sits at that same point. If your chain of inequalities cannot be simultaneously tight where the problem is tight, you have bounded too crudely and the final constant will come out wrong — which is the Nesbitt failure above, diagnosed one step earlier.

This is also why a solution that uses two inequalities has to check both equality conditions at once. Each one is a separate constraint, and satisfying only one of them proves nothing about where the combined bound is attained.

When the answer is both

Go back to where Engel form left Nesbitt: $\frac{(a+b+c)^2}{2(ab+bc+ca)}$. That is not $\frac32$ yet. It is still an expression in the variables, and Cauchy–Schwarz has no more to say about it.

Expand the top: $(a+b+c)^2 = a^2+b^2+c^2 + 2(ab+bc+ca)$. So the fraction is at least $\frac32$ exactly when $a^2+b^2+c^2 \ge ab+bc+ca$ — which is the three-pair square inequality, three applications of $a^2+b^2 \ge 2ab$ added together. That is AM–GM again, in the form $\frac{a+b}{2} \ge \sqrt{ab}$ applied to squares.

That division of labour generalises: Cauchy–Schwarz reduces the expression, AM–GM finishes it. Cauchy–Schwarz is very good at collapsing a shape and almost useless at extracting a number, because the bound it hands back is still symmetric in the variables. Something has to close the gap, and the closer is nearly always AM–GM.

The same order shows up when a product constraint is involved: collapse the fractions with Cauchy–Schwarz first, then spend a constraint like $abc = 1$ with AM–GM. Trying it in the other order usually means bounding a denominator, which is the wrong direction all over again.

Some shapes read as both at once. The lemma $(x+y+z)\left(\frac1x + \frac1y + \frac1z\right) \ge 9$, which Nesbitt's standard solution leans on, is a product of two sums and a sum of terms each paired with its reciprocal. AM–GM on both brackets proves it; Cauchy–Schwarz with $a_i = \sqrt{x_i}$ and $b_i = 1/\sqrt{x_i}$ proves it in one line. Both are correct, and the shape test firing twice is the signal to stop deliberating and write down whichever you can finish faster.

How to train the recognition

Do not solve. Classify. Take twenty inequalities from any source and write two lines on each: which conversion the shape calls for, and where the equality lives. You will get through all twenty in the time one full write-up takes, and classification is the part you are actually short of — the algebra was never the bottleneck.

Lemma teaches these together as classical inequalities rather than as two tricks you meet in unrelated weeks, with the equality case treated as the first step rather than a closing remark. The problem archive has the statements above in full, with difficulty and expected time on each, and the daily problem posts a new one every day in three tiers, no account needed.