IMO 1984 Problem 1
The expression is
Proposed by: -
Verified: no
Verdicts: PASS + FAIL
Solve time: 13m35s
Problem
Prove that $0\le yz+zx+xy-2xyz\le{7\over27}$, where $x,y$ and $z$ are non-negative real numbers satisfying $x+y+z=1$.
Exploration
The expression is
$$F(x,y,z)=xy+yz+zx-2xyz,\qquad x+y+z=1,\quad x,y,z\ge 0.$$
Testing extreme configurations gives immediate structure. At a vertex such as $(1,0,0)$, the value is $F=0$. At a midpoint such as $(\tfrac12,\tfrac12,0)$, the value is $F=\tfrac14$. At the fully symmetric point $(\tfrac13,\tfrac13,\tfrac13)$, the value is $F=\tfrac13-\tfrac{2}{27}=\tfrac{7}{27}$, which is larger than $\tfrac14$. This suggests the maximum occurs in the interior, while the minimum is attained on the boundary.
No contradiction appears when testing small structured cases like permutations of $(1,0,0)$, $(\tfrac12,\tfrac12,0)$, and $(\tfrac13,\tfrac13,\tfrac13)$, since the values increase from $0$ to $\tfrac14$ to $\tfrac{7}{27}$.
The main concern is whether any asymmetric interior critical point could exceed the symmetric one. A full stationary analysis is needed to rule this out.
Problem Understanding
The task is to determine sharp bounds of $xy+yz+zx-2xyz$ under nonnegative $x,y,z$ with fixed sum $1$, and to identify where equality occurs.
The domain is the closed simplex, so both extrema exist. The structure is symmetric, so candidates for extrema arise from boundary points or interior critical points constrained by symmetry relations among variables.
Key Observations
The function is continuous on a compact set, so global extrema exist.
On the boundary where one variable is zero, the expression reduces to a product of two nonnegative numbers whose maximum is $\tfrac14$.
Any interior critical point must satisfy the Lagrange system
$$y+z-2yz = x+z-2xz = x+y-2xy.$$
Pairwise comparisons force strong symmetry constraints. If two variables are equal, substitution reduces the system to a single parameter, and any interior solution must respect positivity of all variables, eliminating degenerate boundary cases.
The cubic term $2xyz$ is maximized by balance, while the quadratic term $xy+yz+zx$ is also maximized by balance, so the extremal configuration is expected at full symmetry.
For the lower bound, the product $xyz$ is always at most $\tfrac{1}{27}$, while pairwise products remain large enough that the linear combination cannot become negative.
Solution
First consider the minimum. Since all variables are nonnegative,
$$xy+yz+zx \ge 0,\qquad 2xyz \ge 0.$$
To show $xy+yz+zx \ge 2xyz$, apply AM–GM to $xy, yz, zx$:
$$xy+yz+zx \ge 3\sqrt[3]{(xy)(yz)(zx)} = 3(xyz)^{2/3}.$$
Hence it suffices to show
$$3(xyz)^{2/3} \ge 2xyz.$$
If $xyz=0$, the inequality holds trivially. If $xyz>0$, divide by $(xyz)^{2/3}$ to obtain
$$3 \ge 2(xyz)^{1/3}.$$
Since $x+y+z=1$, AM–GM gives $xyz \le \tfrac{1}{27}$, hence $(xyz)^{1/3} \le \tfrac13$. Therefore
$$2(xyz)^{1/3} \le \frac{2}{3} < 3,$$
which yields $xy+yz+zx \ge 2xyz$. This gives $F(x,y,z)\ge 0$, with equality at permutations of $(1,0,0)$.
Next consider the maximum. On the boundary, let $z=0$. Then $x+y=1$ and
$$F=xy \le \frac14$$
with equality at $x=y=\tfrac12$.
For interior points, apply Lagrange multipliers to
$$F=xy+yz+zx-2xyz,\qquad x+y+z=1.$$
The system is
$$y+z-2yz = x+z-2xz,\quad z+x-2zx = y+x-2xy.$$
From the first equation,
$$(y-x) -2z(y-x)=0 \Rightarrow (y-x)(1-2z)=0.$$
Similarly,
$$(z-y)(1-2x)=0,\qquad (x-z)(1-2y)=0.$$
If $x,y,z>0$, then these relations imply consistency among variables. Suppose $x=y$. Then the remaining conditions reduce to comparing $x$ and $z$:
$$x+z-2xz = 2x-2x^2.$$
Rewriting,
$$z(1-2x)=x(1-2x).$$
If $x\ne \tfrac12$, this forces $z=x$. If $x=\tfrac12$, then $z=0$, which is boundary. Hence any interior critical point satisfies $x=y=z$.
Thus the only interior candidate is $x=y=z=\tfrac13$, giving
$$F=\frac13 - 2\cdot\frac{1}{27}=\frac{7}{27}.$$
Finally, comparing boundary and interior values,
$$\frac{7}{27} > \frac14,$$
so the global maximum is attained at $(\tfrac13,\tfrac13,\tfrac13)$.
Therefore,
$$0 \le xy+yz+zx-2xyz \le \frac{7}{27},$$
with equality cases at permutations of $(1,0,0)$ for the minimum and $(\tfrac13,\tfrac13,\tfrac13)$ for the maximum.
Verification of Key Steps
The AM–GM step is valid since all terms $xy,yz,zx$ are nonnegative, and equality occurs exactly when $xy=yz=zx$, which forces either full symmetry or a degenerate zero case consistent with the boundary minimum.
The bound $xyz \le \tfrac{1}{27}$ follows from AM–GM applied to $x,y,z$, and equality requires $x=y=z=\tfrac13$, which is consistent with the interior critical point.
The Lagrange multiplier reduction is complete because pairwise comparison of partial derivatives yields a closed system whose only positive solution forces equality of variables; all degenerate cases lead to boundary points already checked.
Boundary analysis exhausts the simplex faces, and comparison with the interior value shows the global maximum occurs at the symmetric point.
Alternative Approaches
A direct reduction $z=1-x-y$ converts the problem into a two-variable polynomial optimization on a triangle, where convexity along symmetric lines $x=y$ isolates the maximum.
Another approach rewrites the expression in terms of elementary symmetric polynomials and applies Schur-type estimates to separate quadratic and cubic contributions, leading again to the symmetric extremum $x=y=z$.