Proving $\dfrac{3}{2}$ Inequality in Acute Triangle $ABC$

[anemone]

In an acute triangle $ABC$, prove that $\dfrac{\cos A}{\cos (B-C)}+\dfrac{\cos B}{\cos (C-A)}+\dfrac{\cos C}{\cos (A-B)}\ge \dfrac{3}{2}$.

 

[jvicens]

I would like to approach this problem by first defining the terms and concepts involved. An acute triangle is a triangle where all three angles are less than 90 degrees. This means that the triangle is not a right triangle or an obtuse triangle. The cosine of an angle is a trigonometric function that relates the length of the adjacent side to the hypotenuse in a right triangle. This can also be extended to any angle in a non-right triangle using the Law of Cosines.

Now, let us consider the given expression: $\dfrac{\cos A}{\cos (B-C)}+\dfrac{\cos B}{\cos (C-A)}+\dfrac{\cos C}{\cos (A-B)}$. We can rewrite this expression as $\dfrac{\cos A}{\cos (B-C)}+\dfrac{\cos B}{\cos (C-B)}+\dfrac{\cos C}{\cos (A-C)}$. This is possible because the cosine function is an even function, meaning that $\cos (-x)=\cos x$.

Next, we can use the fact that in an acute triangle, the sum of the three angles is always less than 180 degrees. Therefore, we can rewrite the expression as $\dfrac{\cos A}{\cos (B-C)}+\dfrac{\cos B}{\cos (C-B)}+\dfrac{\cos C}{\cos (A-C)}=\dfrac{\cos A}{\cos (B-C)}+\dfrac{\cos B}{\cos (C-B)}+\dfrac{\cos C}{\cos (180-A)}$.

Using the Law of Cosines, we know that for an acute triangle, $\cos A, \cos B,$ and $\cos C$ are all positive. This means that we can apply the AM-GM inequality to the three terms in the expression, which states that the arithmetic mean of a set of positive numbers is always greater than or equal to their geometric mean. This can be written as $\dfrac{\dfrac{\cos A}{\cos (B-C)}+\dfrac{\cos B}{\cos (C-B)}+\dfrac{\cos C}{\cos (180-A)}}{3}\ge \sqrt[3]{\dfrac{\cos A}{\cos (B-C)}\cdot \dfrac{\cos B}{\cos (C-B)}\cdot \d