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 Quadrilateral Sine Inequality (Posted on 2008-08-05)
Let A,B,C and D be angles in a quadrilateral. Prove that:

(CSinASinB)+(ASinCSinD)+(D|CosACosB|)+(B|CosCCosD|)    2(pi)

 No Solution Yet Submitted by Jonathan Lindgren No Rating

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 Solution | Comment 1 of 3
Assume largest angle of quadrilateral is less than 180
Apply Cauchy's inequality on LHS to get
(Σ√CSinASinB)² ≤ ΣC*ΣSinASinB -- (#)
ΣC = A+B+C+D = 2*pi --- (##)
(as A,B,C,D are angles of a quadrilateral)
ΣSinASinB = SinASinB+SinCSinD+/-CosACosB+/-CosCCosD
=> ΣSinASinB = (+/- CosACosB+SinASinB) + (+/-CosCCosD+SinCSinD)
1st term can be either Cos(A-B) or - Cos(A+B)
So, (+/- CosACosB+SinASinB) ≤ 1 (since -1 ≤ Cosx ≤ 1)
2nd term can be either Cos(C-D) or -Cos(C+D)
So, (+/-CosCCosD+SinCSinD) ≤ 1 for the same reason
=> ΣSinASinB ≤ 2 --- (###)
Sub eq(###) and eq(##) in eq(#), we get
(Σ√CSinASinB)² ≤ 2*pi*2
=>(Σ√CSinASinB)² ≤ 4*pi
=> (Σ√CSinASinB) ≤ 2√pi

Note: You forgot to consider the case where the largest angle is
greater than 180.

Edited on August 6, 2008, 7:54 am
 Posted by Praneeth on 2008-08-06 07:54:29

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