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Old Nov5-09, 12:40 AM       Last edited by forumfann; Nov5-09 at 02:29 AM..            #1
forumfann

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A basic inequality

Could anyone help me on this,
Is it true that for any given LaTeX Code: r_{1},r_{2},r_{3},r_{4}>0 and LaTeX Code: t_{1},t_{2},t_{3},t_{4}\\in[0,2\\pi) if
LaTeX Code: r_{1}\\left|\\cos(t-t_{1})\\right|+r_{2}\\left|\\cos(t-t_{2})\\right| LaTeX Code: <r_{3}\\left|\\cos(t-t_{3})\\right|+r_{4}\\left|\\cos(t-t_{4})\\right| for all LaTeX Code: t\\in[0,2\\pi)
then LaTeX Code: r_{1}+r_{2}<r_{3}+r_{4} ?

By the way, this is not a homework problem.

Any help will be highly appreciated!
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Old Nov5-09, 03:43 AM       Last edited by Kaimyn; Nov5-09 at 03:52 AM.. Reason: Weird LaTeX code bugs            #2
Kaimyn

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Posts: 42
Re: A basic inequality

Originally Posted by forumfann View Post
Could anyone help me on this,
Is it true that for any given LaTeX Code: r_{1},r_{2},r_{3},r_{4}>0 and LaTeX Code: t_{1},t_{2},t_{3},t_{4}\\in[0,2\\pi) if
LaTeX Code: r_{1}\\left|\\cos(t-t_{1})\\right|+r_{2}\\left|\\cos(t-t_{2})\\right| LaTeX Code: <r_{3}\\left|\\cos(t-t_{3})\\right|+r_{4}\\left|\\cos(t-t_{4})\\right| for all LaTeX Code: t\\in[0,2\\pi)
then LaTeX Code: r_{1}+r_{2}<r_{3}+r_{4} ?

By the way, this is not a homework problem.

Any help will be highly appreciated!
I may be incorrect, but I would say this would be false.

What if LaTeX Code: t-t_{1} and LaTeX Code: t-t_{2} are equal to LaTeX Code: 2pi , LaTeX Code: pi or LaTeX Code: 0 ? Then r1 and r2 can be anything, and don't have to satisfy the inequality!
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Old Nov5-09, 09:27 AM       Last edited by forumfann; Nov5-09 at 11:23 AM..            #3
forumfann

forumfann is Offline:
Posts: 13
Re: A basic inequality

If LaTeX Code: t-t_{1} and LaTeX Code: t-t_{2} are equal to LaTeX Code: 2pi , LaTeX Code: pi or LaTeX Code: 0 ? Then the left hand side of the given inequality is LaTeX Code: r_1+r_2 , which is less than the right hand side of the given inequality that is not larger than LaTeX Code: r_3+r_4 . Thus the claim is automatically true.

I think what makes it possible to be true is "for all LaTeX Code: x\\in[0,2\\pi] ", but I don't know how to prove it.

Again, any suggestion that can lead to the answer to the question will be greatly appreciated.
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Old Nov6-09, 06:40 AM                  #4
Kaimyn

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Posts: 42
Re: A basic inequality

Ahh, sorry, I meant pi/2, meaning cos(pi/2) = 0. Then they do not have to be < r3+ r4

Besides, say they are both equal to pi anyway. Then, r1 and r2 can be greater than r3 and r4, yet still hold true in the first inequaility but not the second.
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