`(cot""A/2+cot""B/2+cot"C/2)/(cot A+cotB+cotC)=((a+b+c)^2)/(a^2+b^2+c^2)`

Answer

To prove the equation

\[ \frac{\cot \frac{A}{2} + \cot \frac{B}{2} + \cot \frac{C}{2}}{\cot A + \cot B + \cot C} = \frac{(a+b+c)^2}{a^2+b^2+c^2} \]

we will start with the left-hand side (LHS) and simplify it step by step.

### Step 1: Express cotangent of half angles
Using the formula for cotangent of half angles, we have:

\[ \cot \frac{A}{2} = \frac{s(s-a)}{\Delta}, \quad \cot \frac{B}{2} = \frac{s(s-b)}{\Delta}, \quad \cot \frac{C}{2} = \frac{s(s-c)}{\Delta} \]

where \( s = \frac{a+b+c}{2} \) (the semi-perimeter) and \( \Delta \) is the area of the triangle.

### Step 2: Substitute into LHS
Substituting these into the LHS gives:

\[ \cot \frac{A}{2} + \cot \frac{B}{2} + \cot \frac{C}{2 = \frac{s(s-a)}{\Delta} + \frac{s(s-b)}{\Delta} + \frac{s(s-c)}{\Delta} \]

Combining these terms yields:

\[ \frac{s[(s-a) + (s-b) + (s-c)]}{\Delta} = \frac{s[3s - (a+b+c)]}{\Delta} = \frac{s[3s - 2s]}{\Delta} = \frac{s^2}{\Delta} \]

### Step 3: Express cotangent of angles
Next, we express \( \cot A, \cot B, \cot C \):

\[ \cot A = \frac{b^2 + c^2 - a^2}{4\Delta}, \quad \cot B = \frac{c^2 + a^2 - b^2}{4\Delta}, \quad \cot C = \frac{a^2 + b^2 - c^2}{4\Delta} \]

### Step 4: Substitute into denominator
Substituting these into the denominator gives:

\[ \cot A + \cot B + \cot C = \frac{(b^2 + c^2 - a^2) + (c^2 + a^2 - b^2) + (a^2 + b^2 - c^2)}{4\Delta} \]

This simplifies to:

\[ \frac{2(a^2 + b^2 + c^2)}{4\Delta} = \frac{a^2 + b^2 + c^2}{2\Delta} \]

### Step 5: Combine LHS
Now, substituting the results from Steps 2 and 4 into the LHS:

\[ \text{LHS} = \frac{\frac{s^2}{\Delta}}{\frac{a^2 + b^2 + c^2}{2\Delta}} = \frac{s^2 \cdot 2}{a^2 + b^2 + c^2} = \frac{2s^2}{a^2 + b^2 + c^2} \]

### Step 6: Substitute \( s \)
Recall that \( s = \frac{a+b+c}{2} \), so:

\[ s^2 = \left(\frac{a+b+c}{2}\right)^2 = \frac{(a+b+c)^2}{4} \]

Thus,

\[ \text{LHS} = \frac{2 \cdot \frac{(a+b+c)^2}{4}}{a^2 + b^2 + c^2} = \frac{(a+b+c)^2}{2(a^2 + b^2 + c^2)} \]

### Step 7: Final simplification
To match the RHS, we need to multiply both sides by 2:

\[ \text{RHS} = \frac{(a+b+c)^2}{a^2 + b^2 + c^2} \]

Thus, we have proved that:

\[ \frac{\cot \frac{A}{2} + \cot \frac{B}{2} + \cot \frac{C}{2}}{\cot A + \cot B + \cot C} = \frac{(a+b+c)^2}{a^2 + b^2 + c^2} \]