Difference between revisions of "1989 OIM Problems/Problem 2"

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== Problem ==
 
== Problem ==
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Let <math>x</math>, <math>y</math>, <math>z</math> three real numbers such that <math>0<x<y<z<\frac{\pi}{2}</math>. Prove the following inequality:
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<cmath>\frac{\pi}{2}+2sin(x)cos(y)+2sin(y)cos(z) > sin(2x)+sin(2y)+sin(2z)</cmath>
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~translated into English by Tomas Diaz. ~orders@tomasdiaz.com
 
~translated into English by Tomas Diaz. ~orders@tomasdiaz.com
Let <math>a</math>, <math>b</math>, and <math>c</math> be the longitudes of the sides of a triangle.  Prove:
 
<cmath>\frac{a-b}{a+b}+\frac{b-c}{b+c}+\frac{c-a}{c+a}<\frac{1}{16}</cmath>
 
  
 
== Solution ==
 
== Solution ==
 
{{solution}}
 
{{solution}}
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== See also ==
 +
https://www.oma.org.ar/enunciados/ibe4.htm

Latest revision as of 13:30, 13 December 2023

Problem

Let $x$, $y$, $z$ three real numbers such that $0<x<y<z<\frac{\pi}{2}$. Prove the following inequality: \[\frac{\pi}{2}+2sin(x)cos(y)+2sin(y)cos(z) > sin(2x)+sin(2y)+sin(2z)\]

~translated into English by Tomas Diaz. ~orders@tomasdiaz.com

Solution

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

See also

https://www.oma.org.ar/enunciados/ibe4.htm