Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "1969 IMO Problems/Problem 6"

(→‎Solution)
(→‎Solution)
Line 13: Line 13:
  
 
<math>\frac{4}{A+B} \le \frac{A+B}{AB}</math>
 
<math>\frac{4}{A+B} \le \frac{A+B}{AB}</math>
 +
 +
<math>\frac{2}{2(A+B)} \le \frac{A+B}{AB}</math>
 +
  
  

Revision as of 22:19, 18 November 2023

Problem

Prove that for all real numbers $x_1, x_2, y_1, y_2, z_1, z_2$, with $x_1 > 0, x_2 > 0, y_1 > 0, y_2 > 0, z_1 > 0, z_2 > 0, x_1y_1 - z_1^2 > 0, x_2y_2 - z_2^2 > 0$, the inequality\[\frac{8}{(x_1 + x_2)(y_1 + y_2) - (z_1 + z_2)^2} \leq \frac{1}{x_1y_1 - z_1^2} + \frac{1}{x_2y_2 - z_2^2}\]is satisfied. Give necessary and sufficient conditions for equality.

Solution

Let $A=x_1y_1 - z_1^2>0$ and $B=x_2y_2 - z_2^2>0$

From AM-GM:

$\sqrt{AB} \le \frac{A+B}{2}$

$4AB \le (A+B)^2$

$\frac{4}{A+B} \le \frac{A+B}{AB}$

$\frac{2}{2(A+B)} \le \frac{A+B}{AB}$


This problem needs a solution. If you have a solution for it, please help us out by adding it.

See Also

1969 IMO (Problems) • Resources
Preceded by
Problem 5 		1 • 2 • 3 • 4 • 5 • 6 Followed by
Last Question
All IMO Problems and Solutions
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1969_IMO_Problems/Problem_6&oldid=204432"