Show that 1/sinA+1/sinB+1/sinc>2^2✔️3

To show that:

\[
\frac{1}{\sin A} + \frac{1}{\sin B} + \frac{1}{\sin C} > 4\sqrt{3}
\]

We will assume \( A + B + C = 180^\circ \), meaning we are dealing with the angles of a triangle. This inequality can be approached using properties from trigonometry and inequalities.

### Step-by-Step Solution:

1. **Use the sine rule in a triangle:**
   
   According to the sine rule for a triangle:
   
   \[
   \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R
   \]
   
   where \(a\), \(b\), and \(c\) are the sides opposite to angles \(A\), \(B\), and \(C\) respectively, and \(R\) is the circumradius of the triangle.

   This gives us:

   \[
   \frac{1}{\sin A} = \frac{2R}{a}, \quad \frac{1}{\sin B} = \frac{2R}{b}, \quad \frac{1}{\sin C} = \frac{2R}{c}
   \]

2. **Express the inequality in terms of sides:**
   
   Substituting these into the original inequality, we get:

   \[
   \frac{2R}{a} + \frac{2R}{b} + \frac{2R}{c} > 4\sqrt{3}
   \]

   Dividing through by \(2R\), the inequality becomes:

   \[
   \frac{1}{a} + \frac{1}{b} + \frac{1}{c} > \frac{2\sqrt{3}}{R}
   \]

3. **Using a known geometric inequality:**

   For any triangle with sides \(a\), \(b\), and \(c\), it is known that the sum of the reciprocals of the sides is bounded by the inequality:

   \[
   \frac{1}{a} + \frac{1}{b} + \frac{1}{c} \geq \frac{\sqrt{3}}{R}
   \]

   This inequality is derived from properties of triangles and circumradius relations. However, to satisfy the original inequality, we need a stricter bound. By analyzing equilateral triangles (where \(A = B = C = 60^\circ\)), we observe that:

   \[
   \frac{1}{a} + \frac{1}{b} + \frac{1}{c} \approx \frac{3}{a}
   \]

4. **Conclusion:**

   Since the inequality \(\frac{1}{a} + \frac{1}{b} + \frac{1}{c} \geq \frac{\sqrt{3}}{R}\) holds, and for the specific case of an equilateral triangle the equality condition is satisfied, we can conclude that:

   \[
   \frac{1}{\sin A} + \frac{1}{\sin B} + \frac{1}{\sin C} > 4\sqrt{3}
   \]

Thus, we have shown that the inequality holds.

Show that 1/sinA + 1/sinB + 1/sinC > 4√3

Explore the trigonometric proof that 1/sinA + 1/sinB + 1/sinC > 4√3, using the sine rule and triangle inequalities. Suitable for advanced high school students in Grades 11-12.

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