# A basic inequality

1. Nov 4, 2009

### 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!

Last edited: Nov 5, 2009
2. Nov 5, 2009

### Kaimyn

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!

Last edited: Nov 5, 2009
3. Nov 5, 2009

### 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.

Last edited: Nov 5, 2009
4. Nov 6, 2009

### 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.