I have posted this inequality to the geometry.puzzle newsgroup. Virtually immediately Antreas Hatzipolakis posted the following:

Alexander Bogomolny asked:

>Is this one known too?
>
>sin(A)*sin(B)/sin(C/2)^2 +
>sin(A)*sin(C)/sin(B/2)^2 +
>sin(B)*sin(C)/sin(A/2)^2 >= 9

No, at least to me. So, next time I see it posted, I will refer to Alex Bogomolny's Inequality :-)

Let's prove it:

In cyclical form reads:

sin(B)*sin(C)/sin(A/2)^2 +
sin(C)*sin(A)/sin(B/2)^2 +
sin(A)*sin(B)/sin(C/2)^2 >= 9

If we de-trigonometrize it, we get:

  s-a     s-b     s-c      9
----- + ----- + ----- >= ---    ( s := semiperimeter)
  a^2     b^2     c^2      4s

==>

-a + b + c    a - b + c    a + b - c         9
------------ + ---------- + ---------- >= ----------
    a^2           b^2          c^2         a + b + c

==>

  b + c       c + a       a + b
(------)^2 + (-----)^2 + (-----)^2 >= 12
    a           b           c


          b + c      b^2 + c^2 + 2bc      4bc
We have: (-----)^2 = ---------------- >= -----
            a              a^2            a^2

etc.

==>

  b + c       c + a       a + b          bc     ca     ab
(----->^2 + (-----)^2 + (-----)^2 >= 4(---- + ---- + ----) =
    a           b           c            a^2    b^2    c^2

        1     1     1
= 4abc(--- + --- + ---) >= 12  QED
       a^3   b^3   c^3

[since for x,y,z > 0 ==> x^3 + y^3 + z^3 >= 3xyz]

Antreas 

                   (s - b)(s - c)    
sin(A/2)sin(A/2) = --------------
                         bc

                   s(s - a)
cos(A/2)cos(A/2) = --------
                      bc

Copyright © 1996-2009 Alexander Bogomolny

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