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

Revision as of 22:20, 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:

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

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

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

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

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Revision as of 22:19, 18 November 2023 (view source) Tomasdiaz (talk \| contribs) (→‎Solution) ← Older edit		Revision as of 22:20, 18 November 2023 (view source) Tomasdiaz (talk \| contribs) Newer edit →
Line 14:		Line 14:
	<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>	+	<math>\frac{8}{2(A+B)} \le \frac{A+B}{AB}</math>

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

Revision as of 22:20, 18 November 2023

Problem

Solution

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