MHB Prove Triangle Inequality: $\sum_{cyc} \sin A$

AI Thread Summary
The discussion centers on proving the inequality for any triangle that states the sum of the sine of angles minus the product of the sine of angles is greater than or equal to the sum of the cubes of the sine of angles. The user expresses uncertainty about using the Rearrangement Inequality in their proof. They acknowledge the inappropriate practice of seeking partial solutions across different forums. A suggested solution is mentioned but not detailed in the discussion. The conversation highlights the complexities involved in proving trigonometric inequalities in triangle geometry.
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Prove, that for any triangle:\[ \sum_{cyc}\sin A - \prod_{cyc}\sin A \ge \sum_{cyc}\sin^3 A \]
 
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I am so sorry, that I have posted a challenge, the solution of which, I am not certain. My problem is the use of the Rearrangement Inequality in the proof. I have asked in the forum: ”Pre-University Math/trigonometry” (http://mathhelpboards.com/trigonometry-12/usage-rearrangement-inequality-trigonometric-expression-20998.html#post95174), and I am aware, that it is bad policy to post a challenge in one forum and ask for a partial solution of it in another on the MHB site. Again, I am very sorry about this. It won´t happen again.

Here is the suggested solution:
Division by $\sin A \sin B \sin C$:

\[\sum_{cyc}\frac{1}{\sin B \sin C}-1 \geq \sum_{cyc}\frac{\sin^2 A}{\sin B \sin C}=\sum_{cyc}\frac{1- \cos^2 A}{\sin B \sin C} \\\\ \Rightarrow \sum_{cyc}\frac{\cos^2 A}{\sin B \sin C} \geq 1\]

Now, here comes the moment, where the Rearrangement Inequality is applied:

\[\sum_{cyc} \frac{\cos^2A}{\sin B\sin C}\geq \sum_{cyc} \frac{\cos B \cos C}{\sin B\sin C} =\sum_{cyc}\cot B \cot C = 1.\]
 
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