Usage of the Rearrangement Inequality in a trigonometric expression

[lfdahl]

In a proof, I encountered the following expressions:

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

My question is concerned with the validity of the inequality.

The inequality is based on the use of the Rearrangement Inequality (RI). Can anyone help me understand exactly how the RI is used here?

Thankyou in advance.

 

[jvicens]

Thank you for bringing up this question. The use of Rearrangement Inequality in this proof is indeed valid and can be explained as follows:

Firstly, let's define the Rearrangement Inequality. It states that if we have two sequences of real numbers \(\{a_1,a_2,...,a_n\}\) and \(\{b_1,b_2,...,b_n\}\) such that \(a_1\geq a_2\geq...\geq a_n\) and \(b_1\geq b_2\geq...\geq b_n\), then the following inequality holds:

\[a_1b_1+a_2b_2+...+a_nb_n\geq a_1b_{\sigma(1)}+a_2b_{\sigma(2)}+...+a_nb_{\sigma(n)},\]

where \(\sigma\) is any permutation of the numbers \(\{1,2,...,n\}\).

Now, let's apply this inequality to the given expressions in the proof. We have the sequences \(\{\cos^2 A, \cos B \cos C, \cot B \cot C\}\) and \(\{\frac{1}{\sin B \sin C},\frac{1}{\sin B \sin C},\frac{1}{\sin B \sin C}\}\). We can see that \(\cos^2 A \geq \cos B \cos C \geq \cot B \cot C\) and \(\frac{1}{\sin B \sin C} \geq \frac{1}{\sin B \sin C} \geq \frac{1}{\sin B \sin C}\). Therefore, by the Rearrangement Inequality, we have:

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

Finally, since \(\sum_{cyc}\cot B \cot C = 1\) (as you have correctly pointed out in your post), we can conclude that:

\[\sum_{cyc}\frac{\cos^2 A}{\sin B \sin C