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I am trying to prove that given $A,B,C,D,E,F \in \big]0,\frac{\pi}{2}\big]$ fixed and $A+C \geq E$ 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?

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  • $\begingroup$ But it is not well defined on the described domain: imagine that $E$ is almost 0. $\endgroup$
    – Fedor Petrov
    Commented Aug 28, 2019 at 15:54
  • $\begingroup$ @FedorPetrov Ah, that is true. I forgot to add another condition. $\endgroup$
    – Loreno Heer
    Commented Aug 28, 2019 at 15:55
  • $\begingroup$ @FedorPetrov I updated the question. $\endgroup$
    – Loreno Heer
    Commented Aug 28, 2019 at 15:58
  • $\begingroup$ I also tried to simplify as: $$\sin(\mu A)\cos(\mu B) + \sin(\mu C)\cos(\mu D) - \sin(\mu \min(A+C, \frac{\pi}{2}))\cos(\mu \arccos(\frac{\sin(A)\cos(B)+\sin(C)\cos(D)}{\sin(\min(A+C, \frac{\pi}{2}))}))$$, but showing that this is smaller than the above and also showing it is still larger than zero does not look like it is easier. $\endgroup$
    – Loreno Heer
    Commented Aug 30, 2019 at 9:53
  • $\begingroup$ may I ask, does this question come from certain geometry problem? $\endgroup$
    – Fedor Petrov
    Commented Sep 3, 2019 at 9:09

1 Answer 1

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I am afraid it is false. Take $F=0$, $A=C$, $E=2A$, $\mu=1/2$. Then we are given $\cos B+\cos D=2\cos A$ and should prove $\cos B/2+\cos D/2\geqslant 2\cos A/2$. But if $\cos B=x$, then $\cos B/2=\sqrt{(1+x)/2}$, this function is concave, thus inverse inequality $\cos B/2+\cos D/2\leqslant 2\cos A/2$ holds.

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  • $\begingroup$ Please note that I specified $A,B,\ldots,F \in ]0,\frac{\pi}{2}]$. In particular $F=0$ is not possible. $\endgroup$
    – Loreno Heer
    Commented Sep 2, 2019 at 10:40
  • $\begingroup$ Yes, but it is certainly not important: take very small $F$ if you do not like zero, and accordingly slightly change $B,D$. $\endgroup$
    – Fedor Petrov
    Commented Sep 2, 2019 at 11:13
  • $\begingroup$ I am not sure I fully understand your example. I see how you derive the first equality, however I believe it should be $\geq$ there. The last inequality does not contradict the required one as there could be equality... $\endgroup$
    – Loreno Heer
    Commented Sep 2, 2019 at 11:31
  • $\begingroup$ It is strict if $B\ne D$. $\endgroup$
    – Fedor Petrov
    Commented Sep 2, 2019 at 12:09
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    $\begingroup$ @LorenoHeer It's generally not a good idea to keep changing the question as responses come in. This usually winds up being frustrating to the community. $\endgroup$
    – Todd Trimble
    Commented Sep 2, 2019 at 14:51

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