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Let x, y and z be real numbers with x, y, z ≥ 0 and x + y + z = 6, and put E = x²y + y²z + z²x…

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Let $x$ , $y$ and $z$ be real numbers with $x, y, z \ge 0$ and $x + y + z = 6$ , and put $E = x^2y + y^2z + z^2x$ . The expression $E$ is cyclic: it keeps its value when the letters are moved round in a cycle, from $(x, y, z)$ to $(y, z, x)$ . But it is not symmetric, because swapping only two of the letters can change its value. So the usual first guess, that the greatest value comes when $x = y = z$ , needs care here. This problem finds the true greatest value of $E$ and every triple that gives it.

Find $E$ when $x = y = z = 2$ .
Suppose $z = 0$ . Find the greatest value of $E$ in this case.
Suppose $y$ lies between $x$ and $z$ , that is, $x \le y \le z$ or $z \le y \le x$ . Prove that $E + xyz \le y(x + z)^2$ , and then that $E + xyz \le 32$ .
Explain why the result of (c) gives $E \le 32$ for all allowed $x$ , $y$ , $z$ , and not only when $y$ lies between the other two. How many triples $(x, y, z)$ give $E = 32$ ?
Put $F = E + xyz$ . Find the greatest value of $F$ , and the number of triples $(x, y, z)$ at which $F$ takes this value.

Count two triples as different when they differ in at least one place, so $(1, 2, 3)$ and $(3, 2, 1)$ are different triples. Write out every step of each proof, and do not assume where the greatest value is.

The worked solution

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Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

Plan. Compare $E$ with $y(x + z)^2$ when $y$ is the middle value, and bound $y(x + z)^2$ by AM-GM. A cyclic shift then covers every case.

(a) $E = 3 \times 2^2 \times 2 = 24$ .

(b) With $z = 0$ we have $E = x^2y$ and $x + y = 6$ . By AM-GM on the three numbers $\dfrac{x}{2}$ , $\dfrac{x}{2}$ , $y$ , whose sum is $6$ : $x^2y = 4 \cdot \dfrac{x}{2} \cdot \dfrac{x}{2} \cdot y \le 4\left(\dfrac{6}{3}\right)^3 = 32$ , with equality when $\dfrac{x}{2} = y$ , that is $x = 4$ , $y = 2$ . So the greatest value is $32$ , which is larger than the value $24$ at the equal point.

(c) Expand: $y(x + z)^2 - (E + xyz) = x^2y + 2xyz + yz^2 - x^2y - y^2z - z^2x - xyz = xyz + yz^2 - y^2z - z^2x$ . This factorises as $z(xy + yz - y^2 - zx) = z(x - y)(y - z)$ . If $y$ lies between $x$ and $z$ , then $x - y$ and $y - z$ are both at most $0$ or both at least $0$ , so their product is at least $0$ , and $z \ge 0$ . Hence $E + xyz \le y(x + z)^2$ . Next, by AM-GM on $y$ , $\dfrac{x + z}{2}$ , $\dfrac{x + z}{2}$ , whose sum is $6$ : $y(x + z)^2 = 4 \cdot y \cdot \dfrac{x + z}{2} \cdot \dfrac{x + z}{2} \le 4 \times 2^3 = 32$ .

(d) Of any three numbers, one lies between the other two. A cyclic shift of the letters, $(x, y, z) \to (y, z, x)$ once or twice, moves that number into the middle place $y$ without changing $E$ or $xyz$ . So $E + xyz \le 32$ always, and $E \le 32 - xyz \le 32$ . For $E = 32$ we need $xyz = 0$ and equality in (c) after the shift: $y = 2$ , $x + z = 4$ and $z(x - y)(y - z) = 0$ . Since $y = 2$ , one of $x$ , $z$ is $0$ . If $z = 0$ then $x = 4$ , giving $(4, 2, 0)$ . If $x = 0$ then $z = 4$ , but then $z(x - y)(y - z) = 4 \times (-2) \times (-2) = 16$ , not $0$ . So, after the shift, only $(4, 2, 0)$ works. Undoing the shift gives the triples $(4, 2, 0)$ , $(0, 4, 2)$ and $(2, 0, 4)$ : there are $3$ of them. Each gives $E = 32$ , for example $16 \times 2 = 32$ for the first.

(e) By (d), $F \le 32$ , with equality exactly when, after the shift, $y = 2$ , $x + z = 4$ and $z(x - y)(y - z) = 0$ . Now $xyz$ need not be $0$ . If $z = 0$ we get $(4, 2, 0)$ . If $x = y$ then $x = 2$ and $z = 2$ . If $y = z$ then $z = 2$ and $x = 2$ . So, after the shift, the triples are $(4, 2, 0)$ and $(2, 2, 2)$ . Undoing the shift gives $(4, 2, 0)$ , $(0, 4, 2)$ , $(2, 0, 4)$ and $(2, 2, 2)$ : the greatest value of $F$ is $32$ , taken at $4$ triples. At $(2, 2, 2)$ , $F = 24 + 8 = 32$ .

Check. On a grid of all triples with steps of $0.05$ , the greatest values of $E$ and $F$ are both $32$ , at exactly the triples above, and fifty thousand random triples never give $F$ above $32$ .

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