1984 IMO Problems/Problem 1

Revision as of 01:04, 19 April 2009 by Isocahedron (talk | contribs) (New page: ==Problem== Let <math>x</math>, <math>y</math>, <math>z</math> be nonnegative real numbers with <math>x + y + z = 1</math>. Show that <math>0 \leq xy+yz+zx-2xyz \leq \frac{7}{27}</math> =...)
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Let $x$, $y$, $z$ be nonnegative real numbers with $x + y + z = 1$. Show that $0 \leq xy+yz+zx-2xyz \leq \frac{7}{27}$


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

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