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.