Nesbitt's inequality: three proofs and how to spot it in disguise

Nobody struggles with Nesbitt's inequality once they have seen a proof. It takes two lines. The problem is that it almost never turns up in its textbook form. It shows up halfway through a harder question, after a substitution, with the letters renamed and the numerators squared, and a student who knows the proof cold still does not see it.

So this is not mainly a proof. It is about recognising the shape, and then choosing which of the three standard proofs fits what is in front of you.

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

$$\frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b} \ge \frac32,$$

with equality exactly when $a = b = c$. That is the classic problem in the archive, and every route below proves it.

What the shape looks like

Three features together mean Nesbitt, and you should check for all three before you commit.

• Each numerator is the letter missing from its denominator. Over $b+c$ sits $a$; over $c+a$ sits $b$. If the numerator shares a letter with its own denominator, it is a different problem.
• The expression does not care about scale. Multiply $a$, $b$, $c$ by the same number and nothing changes: top and bottom both scale. An inequality with that property can be normalised, which is what makes the third proof possible.
• The target is the symmetric value. At $a = b = c$ each fraction is then $\tfrac12$, and three halves is $\tfrac32$. A constant on the right that equals the value at the symmetric point is the problem telling you where equality lives.

Proof one: add 1 to every term

This is the one to reach for when the denominators are plain sums of the other two variables and nothing else is going on.

Adding $1$ to the fraction $\tfrac{a}{b+c}$ turns its numerator into $a+b+c$. Do it to all three and the left side gains $3$, while every numerator becomes the same full sum, which you can pull out. What is left is the full sum times the sum of reciprocals of the three denominators. Call the denominators $x$, $y$, $z$; they add up to twice the full sum. The whole question has become the statement that $(x+y+z)\left(\tfrac1x + \tfrac1y + \tfrac1z\right) \ge 9$, which is AM–GM applied to each bracket. Half of $9$, minus the $3$ you added, is $\tfrac32$.

The move worth keeping is not the $+1$ itself. It is that adding a constant made every numerator identical, and identical numerators factor out. Whenever a cyclic sum has numerators that differ only because each one is missing a different letter, try adding a constant to equalise them.

Proof two: put a square on top

Use this when the numerators are already squares, or when you can see that the problem is going to finish with $a^2 + b^2 + c^2 \ge ab + bc + ca$.

Write $\tfrac{a}{b+c}$ as $\tfrac{a^2}{a(b+c)}$. Same number, but now the numerator is a square, and Cauchy–Schwarz in Engel form applies directly: the sum is at least the square of $a+b+c$ divided by the sum of the three new denominators, which is $2(ab+bc+ca)$. That leaves

$$\frac{(a+b+c)^2}{2(ab+bc+ca)} \ge \frac32,$$

which rearranges to $a^2 + b^2 + c^2 \ge ab + bc + ca$. That is a two-line exercise on its own: it is half the sum of the squares $(a-b)^2$, $(b-c)^2$ and $(c-a)^2$.

The classical inequalities page has the Engel form itself. The reflex to take from here is simpler than the theorem: a fraction with a linear numerator can usually be given a square one by multiplying top and bottom by that numerator.

Proof three: normalise, then work with one variable

This is the route when the expression is scale-free but too messy for either trick above, and it is the one that generalises furthest.

Since scaling does nothing, declare $a + b + c = 1$. Then $b + c = 1 - a$, and each term becomes a function of a single letter, $f(a) = \tfrac{a}{1-a}$. The problem is now: three numbers in $(0, 1)$ add to $1$; show that $f(a) + f(b) + f(c) \ge \tfrac32$.

There are two ways to finish, and both come from the smoothing and normalisation toolkit.

• Tangent line. At the equality point $a = \tfrac13$, the tangent to $f$ is $y = \tfrac94 a - \tfrac14$. The gap between them is $\frac{(3a-1)^2}{4(1-a)}$, which is never negative on $(0, 1)$. Add the three tangent bounds and you get $\tfrac94 - \tfrac34 = \tfrac32$.
• Smoothing. $f$ curves upward on $(0, 1)$, so replacing two unequal values by their average, which keeps the sum at $1$, can only lower $f(a) + f(b)$. Push the variables together and the minimum is where all three are equal.

The tangent line is the one to write in a contest because it is self-contained. Smoothing is the one to think with, because it explains why the answer had to be at $a = b = c$.

The identity behind all three

If you want to know why equality happens only at $a = b = c$, there is an exact formula for how far the sum is above $\tfrac32$:

$$\frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b} - \frac32 = \frac12 \sum \frac{(a-b)^2}{(a+c)(b+c)}.$$

Every term on the right is a square over a positive number, so the sum is never negative, and it is zero only when all three differences vanish. You do not need to memorise this one. It is worth knowing it exists, because if a problem asks you to strengthen Nesbitt, this is the expression you will be bounding.

Where it hides

The point of knowing Nesbitt is spotting it in problems that never print it.

• After a reciprocal substitution. IMO 1995, Problem 2 gives positive $a, b, c$ with $abc = 1$ and asks for a sum of terms $\tfrac{1}{a^3(b+c)}$. Nothing about that looks like Nesbitt. Substitute $x = \tfrac1a$ and the others, use $xyz = 1$, and each term turns into $\tfrac{x^2}{y+z}$: exactly the squared-numerator version. That is a signal to use proof two, and the finish is short from there.
• The squared-numerator version. For positive reals, $\frac{a^2}{b+c} + \frac{b^2}{c+a} + \frac{c^2}{a+b} \ge \frac{a+b+c}{2}$. Unlike Nesbitt itself, this one is sensitive to scale, so the right side carries the variables, and a constraint like $abc = 1$ or $a + b + c = 3$ turns it into a number.
• After normalising. If a problem already says $a + b + c = 1$ and asks about $\frac{a}{1-a} + \frac{b}{1-b} + \frac{c}{1-c}$, that is Nesbitt with the denominators expanded. Undo the normalisation and you will recognise it.
• More variables. With $n$ positive numbers summing to $S$, the sum of $\tfrac{a_i}{S - a_i}$ is at least $\tfrac{n}{n-1}$. Proof one and proof three both carry over almost word for word; that is a good test of whether you understood them.

Two traps

There is no upper bound. For positive reals the sum can be as large as you like: take $a = 1$ and $b$, $c$ tiny, and the first term explodes. If a problem asks for an upper bound, look for a hidden constraint.

In a triangle there is an upper bound. It is $2$. If $a, b, c$ are side lengths, then $a < b + c$, which forces $\tfrac{a}{b+c} < \tfrac{2a}{a+b+c}$, and adding the three gives a sum strictly less than $2$. It is never reached: a triangle squashed flat gets arbitrarily close. Problems that say "sides of a triangle" and ask about this sum want both bounds, $\tfrac32$ from Nesbitt and $2$ from the triangle.

What to do with this

Write the three proofs out once each, from memory, without looking at this page. Then, when you meet any cyclic sum of fractions, ask one question before you start: are the numerators the missing letters? If yes, you already know the answer is at $a = b = c$, and you know three ways in.

The problem archive has the inequalities above with difficulty and expected time on each, and the techniques pages lay out where these tools sit in the ladder. The daily problem is a free way to keep practising between sessions.