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

Revision as of 22:30, 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:

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

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

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

$\frac{8}{2(A+B)} \le \frac{A+B}{AB}$

$\frac{8}{2(A+B)} \le \frac{1}{A}+\frac{1}{B}$ [Equation 1]

$(x_1 + x_2)(y_1 + y_2) - (z_1 + z_2)^2=x_1y_1+x_2y_2+x_1y_2+x_2y_1-z_1^2-2z_1z_2-z_2^2$

$(x_1 + x_2)(y_1 + y_2) - (z_1 + z_2)^2=(A+B)+x_1y_2+x_2y_1-2z_1z_2$

since $x_1y_1>z_1^2$ and $x_2y_2>z_2^2$ ,

then $x_1y_1+x_2y_2-z_1^2-z_2^2 \le x_1y_2+x_2y_1-2z_1z_2$

$(A+B) \le x_1y_2+x_2y_1-2z_1z_2$

$2(A+B) \le x_1y_1+x_2y_2+x_1y_2+x_2y_1-z_1^2-2z_1z_2-z_2^2$

$2(A+B) \le (x_1 + x_2)(y_1 + y_2) - (z_1 + z_2)^2

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

@@ Line 27: / Line 27: @@
 <math>(A+B) \le  x_1y_2+x_2y_1-2z_1z_2</math>
+<math>2(A+B) \le x_1y_1+x_2y_2+x_1y_2+x_2y_1-z_1^2-2z_1z_2-z_2^2</math>
+$2(A+B) \le (x_1 + x_2)(y_1 + y_2) - (z_1 + z_2)^2
 {{solution}}
 == See Also == {{IMO box|year=1969|num-b=5|after=Last Question}}

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

Revision as of 22:30, 18 November 2023

Problem

Solution

See Also