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

• MHB
• anemone
In summary: C}{\cos (180-A)}}$. Finally, we can simplify the right side of the inequality using the fact that$\cos (180-A)=-\cos A$and the Law of Cosines. This gives us$\sqrt[3]{\dfrac{\cos A}{\cos (B-C)}\cdot \dfrac{\cos B}{\cos (C-B)}\cdot \dfrac{\cos C}{-\cos A}}=\sqrt[3]{-\dfrac{\cos A\cdot \cos B\cdot \cos C}{\cos (B-C)\cdot \cos (C-B)}}$. Since the cosine function is a decreasing function in the interval$(0,\pi)$, we anemone Gold Member MHB POTW Director In an acute triangle$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}$. I would like to approach this problem by first defining the terms and concepts involved. An acute triangle is a triangle where all three angles are less than 90 degrees. This means that the triangle is not a right triangle or an obtuse triangle. The cosine of an angle is a trigonometric function that relates the length of the adjacent side to the hypotenuse in a right triangle. This can also be extended to any angle in a non-right triangle using the Law of Cosines. Now, let us consider the given expression:$\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

1. How do you prove the $\dfrac{3}{2}$ inequality in an acute triangle?

The $\dfrac{3}{2}$ inequality in an acute triangle states that the sum of the squares of any two sides of a triangle is greater than the square of the third side. To prove this, we can use the Pythagorean theorem and the fact that the sum of the angles in a triangle is 180 degrees.

2. What is the significance of the $\dfrac{3}{2}$ inequality in an acute triangle?

The $\dfrac{3}{2}$ inequality is significant because it helps us understand the relationship between the sides of an acute triangle. It also allows us to determine if a triangle is acute, as the inequality only holds true for acute triangles.

3. Can the $\dfrac{3}{2}$ inequality be applied to obtuse or right triangles?

No, the $\dfrac{3}{2}$ inequality only applies to acute triangles. In obtuse or right triangles, the sum of the squares of the two shorter sides is equal to the square of the longest side, which violates the inequality.

4. Are there any other ways to prove the $\dfrac{3}{2}$ inequality in an acute triangle?

Yes, there are multiple ways to prove the $\dfrac{3}{2}$ inequality in an acute triangle. One method is to use the Law of Cosines, while another is to use the fact that the area of a triangle is equal to half the product of two sides and the sine of the included angle.

5. How is the $\dfrac{3}{2}$ inequality related to the Triangle Inequality Theorem?

The $\dfrac{3}{2}$ inequality is a special case of the Triangle Inequality Theorem. The Triangle Inequality Theorem states that the sum of any two sides of a triangle must be greater than the third side. The $\dfrac{3}{2}$ inequality is a stronger version of this theorem, as it specifically applies to acute triangles and involves the squares of the sides.

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