Difference between revisions of "1984 IMO Problems/Problem 1"
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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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>. | ||
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+ | == See Also == {{IMO box|year=1984|before=First Question|num-a=2}} |
Latest revision as of 23:45, 29 January 2021
Problem
Let , , be nonnegative real numbers with . Show that
Solution
Note that this inequality is symmetric with x,y and z.
To prove note that implies that at most one of , , or is greater than . Suppose , WLOG. Then, since , implying all terms are positive.
To prove , suppose . Note that since at most one of x,y,z is . Suppose not all of them equals -otherwise, we would be done. This implies and . Thus, define , Then, , , and . After some simplification, since and . If we repeat the process, defining after similar reasoning, we see that .
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 |