Date: Mon, 4 Dec 2000 21:57:10 -0500 (GMT)
From: Alexander Bogomolny
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
P.S. For those who wondered, Antreas arrived at his starting point
using the two inequalities:
(s - b)(s - c)
sin(A/2)sin(A/2) = --------------
bc
s(s - a)
cos(A/2)cos(A/2) = --------
bc
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