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Difference between revisions of "1984 IMO Problems/Problem 1"

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==Solution 2==
 
==Solution 2==
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By the method of Lagrangian multipliers. Let <math>f(x,y,z) = xy + yz + zx - 2xyz</math> and <math>\phi(x,y,z) = x+y+z-1</math>. We will find the local maxima/minima of <math>f</math> over <math>S = \{(x,y,z) \in \mathbb{R}^2 : 0 \leq x,y,z \leq 1\}</math> subject to <math>\phi = 0</math>. Since <math>S</math> is compact every sequence in <math>S</math> has a convergent subsequence. Hence the infimum and supremum of <math>f</math> in <math>S</math> subject to <math>\phi = 0</math> are identifiable with local maxima/minima on the surface <math>x+y+z=1, x,y,z \geq 0</math>. So the method of Lagrangian multipliers will detect the infimum/supremum.
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By the method of Lagrangian multipliers. Let <math>f(x,y,z) = xy + yz + zx - 2xyz</math> and <math>\phi(x,y,z) = x+y+z-1</math>. We will find the local maxima/minima of <math>f</math> over <math>S = \{(x,y,z) \in \mathbb{R}^2 : 0 \leq x,y,z \leq 1\}</math> subject to <math>\phi = 0</math>.  
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Let <math>S' = S \cap \{(x,y,z) \in \mathbb{R}^2 : \phi(x,y,z) = 0\}</math>. <math>S'</math> is bounded and the intersection of two closed sets, hence compact. So every sequence in <math>S'</math> has a convergent subsequence. Hence if <math>(a_k)_k \subset S'</math> is such that <math>f(a_k)</math> converges to the infimum or supremum of <math>f</math> in <math>S'</math>, there will be a convergent subsequence <math>a_{k_j} \rightarrow a_\infty \in S'</math> and <math>f(a_\infty) = \lim_{j \rightarrow \infty}f(a_{k_j})</math> by the continuity of <math>f</math>. Hence the infimum/supremum of <math>f</math> over <math>S'</math> will also be a local minima/maxima of <math>f</math> over <math>S'</math>.  
  
 
We must solve <math>\nabla f - \lambda \nabla \phi = 0</math>. This is equivalent to  
 
We must solve <math>\nabla f - \lambda \nabla \phi = 0</math>. This is equivalent to  

Latest revision as of 12:32, 10 July 2025

Problem

Let $x$, $y$, $z$ be nonnegative real numbers with $x + y + z = 1$. Show that $0 \leq xy+yz+zx-2xyz \leq \frac{7}{27}$

Solution 1

Note that this inequality is symmetric with x,y and z.

To prove \[xy+yz+zx-2xyz\geq 0\] note that $x+y+z=1$ implies that at most one of $x$, $y$, or $z$ is greater than $\frac{1}{2}$. Suppose $x \leq \frac{1}{2}$, WLOG. Then, $xy+yz+zx-2xyz=yz(1-2x)+xy+zx\geq 0$ since $(1-2x)\geq 0$, implying all terms are positive.

To prove $xy+yz+zx-2xyz \leq \frac{7}{27}$, suppose $x \leq y \leq z$. Note that $x \leq y \leq \frac{1}{2}$ since at most one of x,y,z is $\frac{1}{2}$. Suppose not all of them equals $\frac{1}{3}$-otherwise, we would be done. This implies $x \leq \frac{1}{3}$ and $z \geq \frac{1}{3}$. Thus, define \[x' =\frac{1}{3}\], \[y'=y\] \[z'=x+y-\frac{1}{3}\] \[\epsilon = \frac{1}{3}-x\] Then, $x'=x+\epsilon$, $z'=z-\epsilon$, and $x'+y'+z'=1$. After some simplification, \[x'y'+y'z'+z'x'-2x'y'z'=xy+yz+zx-2xyz+(1-2y)(z-x-\epsilon)>xy+yz+zx-2xyz\] since $1-2y>0$ and $z-x-\epsilon=z-\frac{1}{3}>0$. If we repeat the process, defining \[x'' =x'=\frac{1}{3}\] \[y''=\frac{1}{3}\] \[z''=z'+y'-\frac{1}{3}=\frac{1}{3}\] after similar reasoning, we see that \[xy+yz+zx-2xyz\leq x'y'+y'z'+z'x'-2x'y'z' \leq x''y''+y''z''+z''x''-2x''y''z''=\frac{7}{27}\].

Solution 2

By the method of Lagrangian multipliers. Let $f(x,y,z) = xy + yz + zx - 2xyz$ and $\phi(x,y,z) = x+y+z-1$. We will find the local maxima/minima of $f$ over $S = \{(x,y,z) \in \mathbb{R}^2 : 0 \leq x,y,z \leq 1\}$ subject to $\phi = 0$.

Let $S' = S \cap \{(x,y,z) \in \mathbb{R}^2 : \phi(x,y,z) = 0\}$. $S'$ is bounded and the intersection of two closed sets, hence compact. So every sequence in $S'$ has a convergent subsequence. Hence if $(a_k)_k \subset S'$ is such that $f(a_k)$ converges to the infimum or supremum of $f$ in $S'$, there will be a convergent subsequence $a_{k_j} \rightarrow a_\infty \in S'$ and $f(a_\infty) = \lim_{j \rightarrow \infty}f(a_{k_j})$ by the continuity of $f$. Hence the infimum/supremum of $f$ over $S'$ will also be a local minima/maxima of $f$ over $S'$.

We must solve $\nabla f - \lambda \nabla \phi = 0$. This is equivalent to \begin{align} x+y - 2xy &= -2uv + 1/2 = \lambda \\ y+z - 2yz &= -2vw + 1/2 = \lambda \\ z+x - 2zx &= -2wu + 1/2 = \lambda \end{align} where $u = 1/2 - x, v = 1/2 - y, w = 1/2 - z$. If $\lambda = 1/2$ then $uv = vw = wu = 0$. WLOG $u= 0$ and we have $vw = 0$. Then WLOG $v = 0$. These imply $x = y = 1/2$. Then $z = 0$ since $x+y+z = 1$. We get $f(1/2,1/2,0) = 1/4 < 7/27$.

If $\lambda \neq 1/2$ then letting $\lambda' = 1/4 - \lambda/2 \neq 0$ one gets $uv = vw = wu = \lambda'$ which imply $u=v=w$ since $u,v,w \neq 0$. This implies $x=y=z=1/3$ since $x+y+z=1$. And $f(1/3,1/3,1/3) = 7/27$.

Now to check the boundary of $0 \leq x,y,z \leq 1$. WLOG we must consider the cases $z=1$ and $z=0$. If $z=1$ then $x=y=0$ by $x+y+z=1$ so $f(x,y,z) = 0$. If $z = 0$ substituting $y=1-x$ in $f(x,y,0)$ yields \[f(x,1-x,0) = x(1-x) = -(1/2-x)^2+1/4.\] which is between $0$ and $1/4<7/27$ since $0 \leq x \leq 1$. Considering all the values found we find $0 \leq f(x,y,z) \leq 7/27$.

~not_detriti

Video Solution

https://youtu.be/6pI2UoT8AqM

Video Solution

https://youtu.be/U8R86eT_aUo

See Also

1984 IMO (Problems) • Resources
Preceded by
First Question 		1 • 2 • 3 • 4 • 5 • 6 Followed by
Problem 2
All IMO Problems and Solutions
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