Difference between revisions of "1977 USAMO Problems/Problem 5"
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== Solution == | == Solution == | ||
− | {{ | + | By applying the Cauchy-Schwarz Inequality in the form <math>((\sqrt{a})^2 +(\sqrt{b})^2 +(\sqrt{c})^2 +(\sqrt{d})^2 +(\sqrt{e})^2) \left( \left( \dfrac{1}{(\sqrt{a})^2} \right) + \left( \dfrac{1}{(\sqrt{b})^2} \right) + \left( \dfrac{1}{(\sqrt{c})^2} \right) + \left( \dfrac{1}{(\sqrt{d})^2} \right) + \left( \dfrac{1}{(\sqrt{e})^2} \right) \right) \le \left( \dfrac{\sqrt{a}}{\sqrt{a}} + \dfrac{\sqrt{b}}{\sqrt{b}} + \dfrac{\sqrt{c}}{\sqrt{c}} + \dfrac{\sqrt{d}}{\sqrt{d}} + \dfrac{\sqrt{e}}{\sqrt{e}} \right)^2 = (5)^2 = 25</math>, we can easily reduce the given inequality to <math>0 \le 6\left(\sqrt{\frac {p}{q}} - \sqrt {\frac{q}{p}}\right)^2</math>, which is true by the Trivial Inequality. We see that equality is achieved when <math>\left(\sqrt{\frac {p}{q}} - \sqrt {\frac{q}{p}}\right)^2 = 0 \rightarrow \left(\sqrt{\frac {p}{q}} - \sqrt {\frac{q}{p}}\right)=0 \rightarrow \sqrt{\frac {p}{q}}=\sqrt {\frac{q}{p}}</math>, which is achieved when <math>p=q</math>. |
== See Also == | == See Also == |
Revision as of 15:47, 17 June 2015
Problem
If are positive numbers bounded by and , i.e, if they lie in , prove that and determine when there is equality.
Solution
By applying the Cauchy-Schwarz Inequality in the form , we can easily reduce the given inequality to , which is true by the Trivial Inequality. We see that equality is achieved when , which is achieved when .
See Also
1977 USAMO (Problems • Resources) | ||
Preceded by Problem 4 |
Followed by Last Question | |
1 • 2 • 3 • 4 • 5 | ||
All USAMO Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.