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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| + | ==Video Solution== | ||
| + | https://youtu.be/6pI2UoT8AqM | ||
| + | |||
| + | ==Video Solution== | ||
| + | https://youtu.be/U8R86eT_aUo | ||
| + | |||
| + | == See Also == {{IMO box|year=1984|before=First Question|num-a=2}} | ||
Latest revision as of 09:51, 29 June 2022
Problem
Let 
, 
, 
be nonnegative real numbers with 
. Show that
Solution
Note that this inequality is symmetric with x,y and z.
To prove ![\[xy+yz+zx-2xyz\geq 0\]](http://latex.artofproblemsolving.com/8/1/7/817b092a4db75978fc56d08f7bdefd84fb67de60.png)
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 ![\[x' =\frac{1}{3}\]](http://latex.artofproblemsolving.com/c/d/0/cd05886b12da69917131b4af2302a667696d24c2.png)
, ![\[y'=y\]](http://latex.artofproblemsolving.com/b/8/4/b84e769b277b4a168488cb6df408c832d25fd937.png)
![\[z'=x+y-\frac{1}{3}\]](http://latex.artofproblemsolving.com/d/a/a/daa9adad683889f484aaead68d3e6fbd11d3cdfa.png)
![\[\epsilon = \frac{1}{3}-x\]](http://latex.artofproblemsolving.com/1/8/8/188d1a54a162fd5a8f695b33bbcc308013dea3d9.png)
Then, 
, 
, and 
. After some simplification, ![\[x'y'+y'z'+z'x'-2x'y'z'=xy+yz+zx-2xyz+(1-2y)(z-x-\epsilon)>xy+yz+zx-2xyz\]](http://latex.artofproblemsolving.com/7/6/7/767ac5b793687b4181203e44ec8db94a173bf0c2.png)
since 
and 
. If we repeat the process, defining ![\[x'' =x'=\frac{1}{3}\]](http://latex.artofproblemsolving.com/c/5/4/c5441118355611c1b907a7194f9eeb603a012736.png)
![\[y''=\frac{1}{3}\]](http://latex.artofproblemsolving.com/c/5/a/c5a75448067e4c9c362d1cb6f0697b46ad9b22c5.png)
![\[z''=z'+y'-\frac{1}{3}=\frac{1}{3}\]](http://latex.artofproblemsolving.com/4/8/d/48df0d5d942c6ad77c7911b4d6fbf3237e546524.png)
after similar reasoning, we see that ![\[xy+yz+zx-2xyz\leq x'y'+y'z'+z'x'-2x'y'z' \leq x''y''+y''z''+z''x''-2x''y''z''=\frac{7}{27}\]](http://latex.artofproblemsolving.com/1/e/e/1eefb0d900418ce5fedfbccaee2c59eba4aacee4.png)
.
Video Solution
Video Solution
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 | ||
