Subject: Re: Trigonometric inequality
Date: Mon, 10 Nov 1997 21:48:14 -0500
From: Alex Bogomolny

Francois, hello.

I just forgot to update the CTK exchange page. The solution I came up with depends on the fact that the arithmetic mean is never less than the geometric mean.

Multiply the three terms on the left and use sine of the double angle formula to reduce the product to

(*) 64(cos(A/2)*cos(B/2)*cos(C/2))2.

There is a known formula

p = 4Rcos(A/2)*cos(B/2)*cos(C/2)

where p is the semiperemeter and R is the radius of the circumscribed circle.

This reduces (*) to 4p2/R2. Now, if you can prove that this expression attains its minimum for the equilateral triangle you are done. But 4p2/R2 = 27 for the equilateral triangle. Which proves the inequality.

Best regards,
Alexander Bogomolny

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