# $\sin(\mu A)\cos(\mu B) + \sin(\mu C)\cos(\mu D) - \sin(\mu E)\cos(\mu F) \geq 0$ - Version 2.0

I recently asked this question here Inequality involving sine and cosine It turned out that with the conditions I required there, the inequality does not hold. I tried to add extra conditions now, therefore I made a new question:

I am trying to prove that given $$A,B,C,D,E,F \in \big]0,\frac{\pi}{2}\big]$$ fixed and $$A+C \geq E$$, $$B+D \geq E$$, $$A+D≥F$$ and $$B+C≥F$$ and the following equation holds for $$\mu = 1$$:

$$\sin(\mu A)\cos(\mu B) + \sin(\mu C)\cos(\mu D) - \sin(\mu E)\cos(\mu F) \geq 0$$ then it holds for all $$\mu\in [0,1]$$.

Note obviously if $$B \leq F$$ and $$D \leq F$$ the statement is trivial to prove.

I checked numerically and it holds, however whatever I tried to prove it did not work.

Any ideas?

I would like to find a proof of the above. The conditions above should hopefully be sufficient now as far as I can tell.

Edit: As pointed out in the comments, with the four conditions $$A+C \geq E$$, $$B+D \geq E$$, $$A+D≥F$$ and $$B+C≥F$$ the statement should hold. However I could not prove it yet. With only 3 of the conditions counterexamples can be found.

• $A = 1/6, B = 1/5, C=1/3, D=3/2, E=1/2$ and $F = 19/16$. We have $f(1) > 0$ and $f(1/2) < 0$? – River Li Sep 11 '19 at 10:35
• @River Li if that is true then I also need to require $A+D \geq F$ and $B+C \geq F$ which is unfortunate and will make the whole thing more complicated. But I will try to prove it with those conditions. – Loreno Heer Sep 12 '19 at 7:13
• It seems that any three of these four are not enough. With four, I did not find counterexample. – River Li Sep 12 '19 at 9:55