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

(Problem)
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==Solution==
 
==Solution==
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Let <math>A=x_1y_1 - z_1^2>0</math> and <math>B=x_2y_2 - z_2^2>0</math>
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From AM-GM:
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<math>\sqrt{AB} \le \frac{A+B}{2}</math>
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<math>4AB \ (A+B)^2</math>
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{{solution}}
 
{{solution}}
  
 
== See Also == {{IMO box|year=1969|num-b=5|after=Last Question}}
 
== See Also == {{IMO box|year=1969|num-b=5|after=Last Question}}

Revision as of 23:16, 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:

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

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


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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