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Difference between revisions of "1984 IMO Problems/Problem 1"
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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>.
 
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>.
 +
 +
==Video Solution==
 +
https://youtu.be/6pI2UoT8AqM
  
 
== See Also == {{IMO box|year=1984|before=First Question|num-a=2}}
 
== See Also == {{IMO box|year=1984|before=First Question|num-a=2}}

Revision as of 06:27, 3 October 2021

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

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 \[xy+yz+zx-2xyz\leq x'y'+y'z'+z'x'-2x'y'z' \leq x''y''+y''z''+z''x''-2x''y''z''=\frac{7}{27}\].

Video Solution

https://youtu.be/6pI2UoT8AqM

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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