The side of a triangle inscribed in a given circle subtends angles $a, b,$ and $y$ at the center. The minimum value of the arithmetic mean mean of $ \cos (a+ \frac{\pi}{2}), \cos(b+\frac{\pi}{2})$ and $\cos(y+\frac{\pi}{2}) $ is . . . ?

One thing I noticed here is that either $a + b+y = 2 \pi $ a+B+y = pi

or $a+y = b$ enter image description here.

I used a diagram. I don't know if there is a more rigorous way to prove that. The problem can be solved if$ a + b + y = 2 \pi $ but what if it is the other case?

My book solve this problem using AM GM inequality but all these terms are negative so how can that be valid here , also they forgot the other case???

@Drmathva helped me solved it $ with a+b+c = 2 \pi$ case but what about the other case???

  • $\begingroup$ Could you please provide a picture? I don't really understand what you mean in the first sentence... $\endgroup$ – Dr. Mathva May 5 at 19:02
  • $\begingroup$ @Dr.Mathva I am on mobile,. I currently can't . This app will not allow me. I will still try. $\endgroup$ – user541396 May 5 at 19:04
  • $\begingroup$ drive.google.com/file/d/14xSZ2KXuAL4sS5cThTLUyF1etkrh-B4a/…. $\endgroup$ – user541396 May 5 at 19:09
  • $\begingroup$ Oh, see. So you want to minimize $$\sum\cos\big(a+\frac\pi2\big)$$? $\endgroup$ – Dr. Mathva May 5 at 19:12
  • $\begingroup$ @Dr.Mathva right! $\endgroup$ – user541396 May 5 at 19:14

Case 1: $a+b+c=2\pi$

Observe, first of all, that we can expand $$\cos\big(a+\frac\pi2\big)=-\sin a$$

Thus, we want to maximize $$S=\sin a+\sin b+\sin c$$ under the constraint $a+b+c=2\pi$. Since we want to maximize $S$, we want all of it terms to be positive (this is not as rigorous as it might sound, so prove it!). Thus we can assume that $a,b,c\in (0,\pi)$. In that interval $$f(x)=\sin x\implies f''(x)=-\sin x<0$$ We now apply Jense's inequality $$\sin\bigg(\frac{a+b+c}{3}\bigg)=\frac{\sqrt 3}2\ge\frac{\sin a+\sin b+\sin c}3\iff S\le\frac{3\sqrt 3}2$$ Equality holds iff $a=b=c=\frac{2\pi}3$


The second case doesn't really exist

enter image description here In fact, we didn't consider the angle subtended by $\color{fuchsia}c$ correctly! And we are done!

Addendum 2

Please observe that the proof given by your book is wrong!

Consider for instance $f(x):=x+3$. With the same argument given by your book, $f(x)_{min}$ in $(0, \infty)$ is achieved when $x=3$. But this is wrong!

Of course $$x+3\ge 2\cdot \sqrt{3x}$$

enter image description here

and they are equal when $x=3$. But this doesn't imply that $x=3$ gives the minimum value for $f$...

  • $\begingroup$ Isn't cos negative in second quadrant? $\endgroup$ – user541396 May 5 at 19:20
  • $\begingroup$ Upps... You're right $\endgroup$ – Dr. Mathva May 5 at 19:22
  • $\begingroup$ In your answer you made an assumption that triangle is distributed . But what if it is on the one side of diameter , drive.google.com/file/d/1T1YXVj3eerymrlegCNiV56v1lYDQfpUA/… $\endgroup$ – user541396 May 5 at 19:33
  • $\begingroup$ Done!! @swarnim $\endgroup$ – Dr. Mathva May 7 at 17:16
  • $\begingroup$ How can one be sure Jensen's inequality will not make the same mistake as am-gm??? $\endgroup$ – user541396 May 15 at 21:20

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