The $uvw$ Method

The $uvw$ method is a useful technique for proving inequalities involving symmetric polynomials in three real nonnegative variables. These types of problems are common in contest mathematics. The basic idea is to introduce a specific change of variables that simplifies the original inequality.

Let $a,b,c$ be nonnegative real numbers. The $uvw$ method refers to the following substitutions:

$$\begin{aligned} 3u &= a+b+c \\ 3v^2 &= ab + bc + ca \\ w^3 &= abc. \end{aligned}$$

Note that the right sides of these substitutions are the three elementary symmetric polynomials in $a,b,c,$ so every symmetric polynomial in $a,b,c$ can be written as a polynomial in $u,v^2,w^3.$

The $uvw$ method is commonly used to show that the maximum or minimum value of an expression involving nonnegative real numbers $a,b,c$ is attained when two of the variables are equal or one of the variables is zero. This requires some important facts about the expressions $u,v,w,$ which are given in the following sections.

The first fact about the change of variables is a simple inequality involving $u,v,w.$ This basic fact motivates the particular choice of variables for the method.

(Basic Theorem) If $a,b,c$ are nonnegative and $u,v,w$ are the substitutions in the $uvw$ method, then $u \ge v \ge w,$ with equality holding if and only if $a=b=c$ or (for $v \ge w$) if two of $a,b,c$ are $0.$

First, $$\begin{aligned} 18u^2 - 18v^2 &= 2(a+b+c)^2 - 6(ab+bc+ac) \\ &= 2a^2+2b^2+2c^2-2ab-2bc-2ca \\ &= (a-b)^2+(b-c)^2+(c-a)^2 \\ &\ge 0, \end{aligned}$$ so $u \ge v,$ with equality if and only if $a-b=b-c=c-a=0,$ i.e. $a=b=c.$

Second, $v^2$ is the arithmetic mean of $ab,bc,ca,$ so by AM-GM it is $\ge$ the geometric mean of $ab,bc,ca,$ which is $(abc)^{2/3} = w^2.$ So $v^2\ge w^2,$ so $v \ge w,$ and equality holds if and only if $ab=bc=ca,$ which only happens if $a=b=c$ or two of the variables are $0.$ $_\square$

Here is an easy example that uses only the basic theorem.

Suppose $a,b,c$ are positive real numbers such that $\dfrac1a + \dfrac1b + \dfrac1c = 1.$ Find the minimum value of $(a-1)(b-1)(c-1).$

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After clearing denominators, the constraint becomes $ab+bc+ca = abc,$ or $3v^2 = w^3.$ Now $$(a-1)(b-1)(c-1) = abc - (ab+bc+ca) + (a+b+c) - 1 = w^3 - 3v^2 + 3u - 1 = 3u - 1.$$ Since $u\ge v \ge w,$ it is clear that $uv^2 \ge w^3,$ so $u(3v^2) \ge 3w^3,$ but since $3v^2 =w^3,$ canceling gives $u \ge 3.$ So $3u-1 \ge 8.$

Finally, the value $8$ is attained when $a=b=c=3,$ so $8$ is in fact the minimum value of $(a-1)(b-1)(c-1).$ $_\square$

Using the $uvw$ method in this way is not particularly powerful: it consists of some simplifying substitutions and a translation of two very basic inequalities (the ones used to prove the basic theorem). The real power of the method comes from a general result known as Tejs' theorem, which is stated and proved below. Tejs' theorem is often used to provide a rigorous justification for a common shortcut: it shows that under certain circumstances, the maximum or minimum of a symmetric expression in three nonnegative real variables occurs when two of the variables are equal, or one of the variables is 0. (In fact, the theorem can be used in more complicated cases, in which more types of triples $(a,b,c)$ must also be checked; for an example, see the last section of the wiki.)

When making a change of variables in general, it is often quite important to understand its inverse. In this particular case, passing from the variables $a,b,c$ to $u,v,w$ is relatively easy to understand. But suppose we are given (nonnegative) values of $u,v,w$ and we wish to recover the corresponding $a,b,c.$ This is a more delicate process, and some values of $u,v,w$ may not even correspond to real values of $a,b,c.$ The following lemma gives conditions that are necessary and sufficient for values of $u,v,w$ to correspond to real values of $a,b,c.$

The real numbers $u,v,w$ satisfy the equations $$\begin{aligned} 3u &= a+b+c \\ 3v^2 &= ab + bc + ca \\ w^3 &= abc \end{aligned}$$ for some real numbers $a,b,c$ if and only if $T(u,v,w) \ge 0,$ where $$T(u,v,w) = -4u^3w^3 + 3u^2v^4 + 6uv^2w^3 - 4v^6 - w^6.$$ Moreover, the following holds: $$T(u,v,w) \ge 0 \text{ and } u,v^2,w^3 \text{ are all nonnegative} \Leftrightarrow a,b,c \text{ are all nonnegative.}$$

This is an application of the cubic discriminant and Vieta's formulas. The key fact is that $u,v,w$ satisfy the equations given in the statement if and only if the polynomial $$x^3-3ux^2+3v^2x-w^3 = (x-a)(x-b)(x-c)$$ has three real roots. A cubic polynomial has three real roots if and only if its discriminant $\Delta = (a-b)^2(b-c)^2(c-a)^2$ is nonnegative. Writing $\Delta$ in terms of $u,v,w$ gives $$\Delta = 27\big(-4u^3w^3 + 3u^2v^4 + 6uv^2w^3 - 4v^6 - w^6\big),$$ which is nonnegative if and only if the quantity in parentheses is nonnegative. This quantity is $T(u,v,w).$

As for the statement about nonnegativity, it's clear that if $a,b,c$ are nonnegative, then $u,v^2,w^3$ are, and $T(u,v,w)$ is (because it's a product of squares of real numbers). For the converse, suppose $u,v^2,w^3$ are all nonnegative. Proceed by contradiction: if $a,b,c$ are not all nonnegative, then at least one is negative. Suppose without loss of generality that $a<0.$ Since $a+b+c$ is nonnegative, at least one of them is positive, say without loss of generality that $c > 0.$ Note also that $b+c$ must be positive. Since $abc$ is nonnegative, $b \le 0.$ But then $ab + bc + ca = a(b+c)+bc$ is the sum of a negative number and a number that is $\le 0,$ so it is negative. This is a contradiction, so $a,b,c$ are all nonnegative. $_\square$

This lemma will be useful in the proof of Tejs' theorem below. Also, the lemma motivates some notation which will be used in the rest of this wiki.

A triple $(u,v,w)$ of nonnegative real numbers will be called admissible if there exist nonnegative real numbers $a,b,c$ such that

$$\begin{aligned} 3u &= a+b+c \\ 3v^2 &= ab + bc + ca \\ w^3 &= abc. \end{aligned}$$

So the lemma says that $(u,v,w)$ is admissible if and only if $T(u,v,w) \ge 0.$

The idea behind Tejs' theorem is to write a polynomial expression in $a,b,c$ as an expression in $u,v^2,w^3,$ and then to ask where the maximum or minimum of this expression occurs. In particular if two of the three variables $u,v,w$ are fixed, the expression becomes a polynomial in the third variable, and the maximum/minimum computation becomes a single-variable optimization problem.

The algebra involved can be unpleasant, and the polynomial condition on $u,v,w$ (from the previous section) is unwieldy, so this is not often a practical way to solve an inequality explicitly. The striking conclusion of Tejs' theorem is that under certain conditions, this technique always yields a maximum or minimum in two very specific cases: either two of $a,b,c$ are equal or one of $a,b,c$ is zero. So solving an inequality using Tejs' theorem usually boils down to verifying that the "certain conditions" hold, and then verifying the inequality when $a=b$ or $c=0.$

(Tejs' Theorem)

(1) Suppose $u,v$ are fixed nonnegative real numbers. Let $S_{u,v}$ be the set of nonnegative real numbers $w$ such that $(u,v,w)$ is admissible. Then if $S_{u,v}$ is nonempty, it has both minimum and maximum elements, and these elements correspond to triples $(a,b,c)$ for which at least two of the variables are equal, or (possibly for the minimum element) one of the variables is zero.

(2) Suppose $u,w$ are fixed nonnegative real numbers. Let $S_{u,w}$ be the set of nonnegative real numbers $v$ such that $(u,v,w)$ is admissible. Then if $S_{u,w}$ is nonempty, it has both minimum and maximum elements, and these elements correspond to triples $(a,b,c)$ for which at least two of the variables are equal.

(3) Suppose $v,w$ are fixed nonnegative real numbers, and suppose $w \ne 0.$ Let $S_{v,w}$ be the set of nonnegative real numbers $u$ such that $(u,v,w)$ is admissible. Then if $S_{v,w}$ is nonempty, it has both minimum and maximum elements, and these elements correspond to triples $(a,b,c)$ for which at least two of the variables are equal.

(1) The condition that $(u,v,w)$ is admissible is $T(u,v,w) \ge 0,$ i.e. $$-w^6 + (6uv^2-4u^3)w^3 + (3u^2v^4-4v^6) \ge 0.$$ With $u,v$ fixed, this is a quadratic polynomial in $w^3,$ say $f(w^3).$ The graph of $f$ is a parabola pointing down. If $S_{u,v}$ is empty, there is nothing to prove, so assume it is nonempty. Then we seek the set of nonnegative $w^3$ for which $f(w^3)$ is nonnegative. The set of points $x$ for which $f(x)$ is nonnegative is an interval $[x_0,x_1].$ Note that $x_1 \ge 0$ because otherwise there would be no nonnegative $w^3$ for which $f(w^3)$ was nonnegative, so $S_{u,v}$ would be empty. There are two cases.

If $x_0 \ge 0,$ the endpoints $x_0$ and $x_1$ are the min and max values of $w^3,$ and they correspond to the values of $w^3$ for which $T$ is zero. But $T(u,v,w) = 0$ if and only if $(a-b)^2(b-c)^2(c-a)^2=0,$ which happens if and only if at least two of $a,b,c$ are equal. So the max and min occur at values where two variables are equal.

If $x_0 < 0,$ the minimum value of $S_{u,v}$ is $w=0.$ So $abc=0,$ so one of the variables is zero. The max $x_1$ still corresponds to two of $a,b,c$ being the same, by the same argument as the previous paragraph.

(2) Now we search for nonnegative values of $v$ such that $T(u,v,w) \ge 0.$ This time $T(u,v,w)$ is a cubic polynomial in $v^2,$ say $g(v^2).$ Note that $g(x)$ is negative for large enough $x$ because its lead term is $-4x^3.$ Again, the point is that the set of nonnegative values of $x$ for which $g(x)$ is nonnegative is, if it is nonempty, a finite union of closed intervals, whose endpoints are either $0$ or values of $x$ for which $g(x) = 0.$ The maximum element of this set will be a value of $v^2$ such that $T(u,v,w) =0,$ which corresponds to two of $a,b,c$ being equal, and the minimum will either be another value of $v^2$ for which $T(u,v,w)=0,$ or $v^2=0.$ But if $v^2=0,$ then $ab+bc+ac=0,$ and since $a,b,c$ are nonnegative, this is only possible if $ab=bc=ac=0,$ which only happens if at least two of $a,b,c$ are zero (hence equal).

(3) Now we search for nonnegative values of $u$ such that $T(u,v,w) \ge 0.$ If $w \ne 0,$ this is a cubic in $u,$ whose lead term is $-4w^3,$ which is negative. The same argument as in the previous paragraph applies. (In this case, $u=0$ corresponds to $a+b+c=0,$ which implies that $a=b=c=0.$) $_\square$

The theorem has the following two immediate corollaries, which give tools to solve many problems involving symmetric inequalities in three variables:

(1) Any symmetric polynomial of degree $\le 5$ in nonnegative real variables $a,b,c$ with a global minimum and/or maximum will attain this value at triples $(a,b,c)$ with either two of the variables equal or one of the variables equal to zero.

(2) Let $f$ be a symmetric polynomial of degree $\le 8$ in nonnegative real variables $a,b,c.$ Write $f$ as $Aw^6+Bw^3+C,$ where $A,B,C$ are functions of $u,v^2.$ Then a global minimum and/or maximum of $f,$ if it exists, will be attained at triples $(a,b,c)$ with either two of the variables equal or one of the variables equal to zero, or in the places corresponding to solutions of $2Aw^3+B=0.$

For (1), fix $u,v$ and consider the resulting polynomial in $w^3$; it is either constant or linear. So its extrema occur when $w^3$ is maximized or minimized, which happens only at the special triples $(a,b,c)$ given in part (1) of Tejs' theorem.

For (2), do the same and notice that the resulting polynomial is at most quadratic in $w^3.$ So its extrema will occur either at points where $w^3$ is maximized or minimized (i.e. at the endpoints of the domain) or when the quadratic polynomial is at a critical point. Such critical points correspond to places where $2Aw^3+B=0,$ by elementary calculus. (The point is that this equation might be easier to analyze than the original one, since it has lower degree.) $_\square$

Let $a,b,c$ be nonnegative real numbers such that $a+b+c=1.$ Show that $$1+12abc \ge 4(ab+bc+ca).$$

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Consider the quantity $1+12abc-4(ab+bc+ca).$ This is a symmetric polynomial of degree 3, namely $1+12w^3-\frac43 v^2.$ It's linear in $w^3$ with a positive linear coefficient, so it is minimized when $w^3$ is 0 or when $w^3$ is minimal subject to the admissibility constraint, which happens when two variables are the same or one is zero by Tejs' theorem. So it suffices to show the inequality for those particular cases.

When (wlog) $a=0,$ we get $b+c=1$ and our quantity is $1-4bc.$ When $b,c$ are nonnegative, $b+c=1$ implies $bc \le \frac 14$ (by AM-GM), so $1-4bc \ge 1-4\big(\frac 14\big) = 0.$ Note that this minimum is attained when $b=c=\frac 12.$

When (wlog) $a=b$ we get $2b+c=1$ and our quantity is $1+12b^2c-4(b^2+2bc) = (12b^2-8b)c + (1-4b^2).$ Substituting $c=1-2b$ gives the quantity as $(12b^2-8b)(1-2b)+1-4b^2 = (1-2b)(12b^2-6b+1).$ The second factor is always positive, and the first factor is positive on the interval $\big[0,\frac 12\big],$ which is where $b$ is constrained to lie because $2b+c=1$ and $b,c$ are nonnegative. So the quantity is always $\ge 0$ in this case as well. $_\square$

Let $a,b,c$ be nonnegative real numbers such that $ab+bc+ca=1.$ Find the minimum value of $$\frac1{a+b} + \frac1{b+c} + \frac1{c+a}.$$

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Setting $a=b=1$ and $c=0$ gives $\frac 52.$ We will show that in fact this is the minimum: $$\frac1{a+b} + \frac1{b+c} + \frac1{c+a} \ge \frac 52.$$

To do this, note that this inequality is equivalent to $$\begin{aligned} (b+c)(c+a)+(a+b)(c+a)+(a+b)(b+c) &\ge \frac52 (a+b)(b+c)(c+a), \text{ or} \\ (b+c)(c+a)+(a+b)(c+a)+(a+b)(b+c) &- \frac52 (a+b)(b+c)(c+a) \ge 0. \end{aligned}$$ The left side is a symmetric polynomial in $a,b,c$ of degree $3.$ In terms of $u,v,w,$ it is $$9u^2+3v^2+\frac52(w^3-9uv^2),$$ which equals $\frac52 w^3 + 9u^2-\frac{15}2 u + 1$ in our case (since $3v^2=1$). For fixed $u$ this is a linear polynomial in $w^3,$ which will attain its minimum value either when $w^3=0$ or when $w^3$ is minimized subject to the admissibility constraint, which happens when two of the variables are equal or one of the variables is zero. So it is enough to look for the minimum value of this expression when two of the variables are equal or one of the variables is zero.

When one of the variables is zero, without loss of generality $a=0$ and $bc=1.$ The expression becomes $9u^2-\frac{15}2 u + 1,$ where $3u=b+c,$ and $bc=1,$ so this factors as $\big(3u-\frac 12\big)(3u-2) = \big(b+c-\frac 12\big)(b+c-2).$ But when $bc=1,$ clearly $b+c \ge 2$ (by AM-GM), so the expression is always nonnegative, and attains its minimum value of $0$ when $b=c=1.$

When two of the variables are equal, without loss of generality, $a=b.$ Then rewriting in terms of $b,c$ (and using the factorization of $9u^2-\frac{15}2 u + 1$ in the previous paragraph) gives the constraint as $b^2+2bc=1$ and the expression as $$\dfrac52 b^2 c + \left(2b+c-\frac12\right)(2b+c-2),$$ and the goal is to show that this is always $\ge 0.$ Making the substitution $c = \frac{1-b^2}{2b}$ gives $$\frac{-5b^5 + 9b^4 - 10b^3 + 10b^2 - 5b + 1}{4b^2}$$ which factors as $$\frac{(1-b)(b^2+1)(5b^2-4b+1)}{4b^2},$$ and each of the three factors on top (as well as the denominator) are always nonnegative for $b \in [0,1]$ $($this interval is forced by the constraint $b^2+2bc=1)$, so the expression is $\ge 0.$ $_\square$

As the previous example shows, the $uvw$ method may still require quite a lot of computation after applying Tejs' theorem.

It is tempting to conclude that Tejs' theorem shows that any inequality involving a symmetric polynomial in three variables need only be checked when two of the variables are equal or one of the variables is zero. This is wrong, as the following example shows:

Let $a,b,c$ be nonnegative real numbers with $a+b+c=12$ and $a^2+b^2+c^2=54.$ Find the maximum value of $102abc - a^2b^2c^2.$

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Suppose we check only cases where two of the variables are equal or one variable is zero. If $a=0$ we get $b+c=12$ and $b^2+c^2=54,$ so $b+c=12$ and $bc=45,$ which is impossible by AM-GM. If $a=b$ we get $2a+c=12$ and $2a^2+c^2=54,$ so $2a^2+(12-2a)^2 = 54,$ so $6a^2-48a+90 = 0,$ which factors as $6(a-3)(a-5)=0.$ This leads to the two solutions $(3,3,6)$ and $(5,5,2).$ In the first case, $abc = 54,$ so $102abc-a^2b^2c^2 = 2592,$ and in the second case, $abc=50,$ so $102abc-a^2b^2c^2=2600.$ So we might erroneously conclude that the minimum is $2592$ and the maximum is $2600,$ but this is incorrect.

In fact it is not hard to see that the possible values of $abc$ given the constraints are in the closed interval $[50,54],$ by applying Tejs' theorem, or by looking at the graph of the cubic function $y = x^3-12x^2+45x - d$ for various values of $d,$ and concluding that it crosses the $x$-axis three times if and only if $d \in [50,54].$ However, the quantity $102abc-a^2b^2c^2$ is clearly maximized when $abc = 51,$ which occurs at neither of the endpoints of the interval. The maximum value is therefore $2601,$ occurring when $a,b,c$ are the three real roots of the polynomial $x^3-12x^2+45x-51.$

This is the situation described in corollary (2) of Tejs' theorem; the maximum of the expression $102w^3-w^6$ occurs either when two of the variables are equal or when the quadratic polynomial in $w^3$ has a critical point, i.e. $-2w^3+102=0,$ or $w^3=51.$ Polynomials in $w^3$ of larger degree will have more critical points, which will be more difficult to compute and will need to be checked more carefully. (This case is extremely easy even compared to the general degree-6 case, because the quadratic in $w^3$ might have coefficients involving $u$ and $v$ instead of just rational numbers.) This is why the $uvw$ method is best for symmetric polynomials of low degree. $_\square$