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

Revision as of 00:04, 19 April 2009

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

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

Revision as of 00:04, 19 April 2009

Problem

Solution