1984 IMO Problems/Problem 1: Difference between revisions

Revision as of 02:14, 9 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 $\phi = 0$ . Since $S$ is compact every convergent sequence in $S$ has a convergent subsequence. Hence the infimum and supremum of $f$ in $S$ subject to $\phi = 0$ are identifiable with local maxima/minima on the surface $x+y+z=1, x,y,z \geq 0$ . So the method of Lagrangian multipliers will detect the infimum/supremum.

We must solve $\nabla f - \lambda \nabla \phi = 0$ . This is equivalent to \begin{align} x+y - 2xy - \lambda &= -2uv + 1/2 - \lambda = 0 \\ y+z - 2yz - \lambda &= -2vw + 1/2 - \lambda = 0 \\ z+x - 2zx - \lambda &= -2wu + 1/2 - \lambda = 0 \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 $x = y = 1/2$ and $z = 0$ 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 $f(x,y,z) = 0$ since $x+y+z = 1$ . 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 $0 \leq x \leq 1$ . The other case of the boundary (e.g. $x=1$ ) implies $f(x,y,z) = 0$ . Considering all the values found we find $0 \leq f(x,y,z) \leq 7/27$ .

~detriti

Video Solution

https://youtu.be/6pI2UoT8AqM

Video Solution

https://youtu.be/U8R86eT_aUo

@@ Line 3: / Line 3: @@
 <math>0 \leq xy+yz+zx-2xyz \leq \frac{7}{27}</math>
-==Solution==
+==Solution 1==
 Note that this inequality is symmetric with x,y and z.
@@ Line 9: / Line 9: @@
 To prove <math>xy+yz+zx-2xyz \leq \frac{7}{27}</math>, suppose <math>x \leq y \leq z</math>. Note that <math>x \leq y \leq \frac{1}{2}</math> since at most one of x,y,z is <math>\frac{1}{2}</math>.  Suppose not all of them equals <math>\frac{1}{3}</math>-otherwise, we would be done. This implies <math>x \leq \frac{1}{3}</math> and <math>z \geq \frac{1}{3}</math>. Thus, define <cmath>x' =\frac{1}{3}</cmath>, <cmath>y'=y</cmath> <cmath>z'=x+y-\frac{1}{3}</cmath> <cmath>\epsilon = \frac{1}{3}-x</cmath> Then, <math>x'=x+\epsilon</math>, <math>z'=z-\epsilon</math>, and <math>x'+y'+z'=1</math>. After some simplification, <cmath>x'y'+y'z'+z'x'-2x'y'z'=xy+yz+zx-2xyz+(1-2y)(z-x-\epsilon)>xy+yz+zx-2xyz</cmath> since <math>1-2y>0</math> and <math>z-x-\epsilon=z-\frac{1}{3}>0</math>. If we repeat the process, defining <cmath>x'' =x'=\frac{1}{3}</cmath> <cmath>y''=\frac{1}{3}</cmath> <cmath>z''=z'+y'-\frac{1}{3}=\frac{1}{3}</cmath> after similar reasoning, we see that <cmath>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}</cmath>.
+==Solution 2==
+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 convergent 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.
+We must solve <math>\nabla f - \lambda \nabla \phi = 0</math>. This is equivalent to
+\begin{align}
+x+y - 2xy - \lambda &= -2uv + 1/2 - \lambda = 0 \\
+y+z - 2yz - \lambda &= -2vw + 1/2 - \lambda = 0 \\
+z+x - 2zx - \lambda &= -2wu + 1/2 - \lambda = 0
+\end{align}
+where <math>u = 1/2 - x, v = 1/2 - y, w = 1/2 - z</math>. If <math>\lambda = 1/2</math> then <math>uv = vw = wu = 0</math>. WLOG <math>u= 0</math> and we have <math>vw = 0</math>.
+Then WLOG <math>v = 0</math>. These imply <math>x = y = 1/2</math> and <math>z = 0</math> since <math>x+y+z = 1</math>. We get <math>f(1/2,1/2,0) = 1/4 < 7/27</math>.
+If <math>\lambda \neq 1/2</math> then letting <math>\lambda' = 1/4 - \lambda/2 \neq 0</math> one gets <math>uv = vw = wu = \lambda'</math> which imply <math>u=v=w</math> since <math>u,v,w \neq 0</math>. This implies <math>x=y=z=1/3</math> since <math>x+y+z=1</math>. And <math>f(1/3,1/3,1/3) = 7/27</math>.
+Now to check the boundary of <math>0 \leq x,y,z \leq 1</math>. WLOG we must consider the cases <math>z=1</math> and <math>z=0</math>. If <math>z=1</math> then <math>f(x,y,z) = 0</math> since <math>x+y+z = 1</math>. If <math>z = 0</math> substituting <math>y=1-x</math> in <math>f(x,y,0)</math> yields
+<cmath>
+f(x,1-x,0) = x(1-x) = -(1/2-x)^2+1/4.
+</cmath>
+which is between <math>0</math> and <math>1/4<7/27</math> since <math>0 \leq x \leq 1</math>. The other case of the boundary (e.g. <math>x=1</math>) implies <math>f(x,y,z) = 0</math>. Considering all the values found we find <math>0 \leq f(x,y,z) \leq 7/27</math>.
+~detriti
 ==Video Solution==

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1984 IMO Problems/Problem 1: Difference between revisions

Revision as of 02:14, 9 July 2025

Contents

Problem

Solution 1

Solution 2

Video Solution

Video Solution

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