Solving Basic Inequality: r1, r2, r3, r4 >0 & t1, t2, t3, t4 in [0, 2π)

[forumfann]

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

By the way, this is not a homework problem.

Any help will be highly appreciated!

 

[Kaimyn]

Could anyone help me on this,
Is it true that for any given r_{1},r_{2},r_{3},r_{4}>0 and t_{1},t_{2},t_{3},t_{4}\in[0,2\pi) if
r_{1}\left|\cos(t-t_{1})\right|+r_{2}\left|\cos(t-t_{2})\right|<r_{3}\left|\cos(t-t_{3})\right|+r_{4}\left|\cos(t-t_{4})\right| for all t\in[0,2\pi)
then 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 t-t_{1} and t-t_{2} are equal to 2pi, pi or 0? Then r₁ and r₂ can be anything, and don't have to satisfy the inequality!

 

[forumfann]

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

I think what makes it possible to be true is "for all 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.

 

[Kaimyn]

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.