To prove that if \(\frac{1}{a(b+c)}, \frac{1}{b(c+a)}, \frac{1}{c(a+b)}\) are in Harmonic Progression (H.P.), then \(a, b, c\) are also in H.P., we will follow these steps:
### Step 1: Understanding H.P.
Recall that three numbers \(x, y, z\) are in H.P. if their reciprocals \(\frac{1}{x}, \frac{1}{y}, \frac{1}{z}\) are in Arithmetic Progression (A.P.).
### Step 2: Set up the condition for H.P.
Given that \(\frac{1}{a(b+c)}, \frac{1}{b(c+a)}, \frac{1}{c(a+b)}\) are in H.P., we can write:
\[
\frac{1}{a(b+c)}, \frac{1}{b(c+a)}, \frac{1}{c(a+b)}
\]
are in H.P. implies:
\[
2 \cdot \frac{1}{b(c+a)} = \frac{1}{a(b+c)} + \frac{1}{c(a+b)}
\]
### Step 3: Cross-multiply to eliminate fractions
Cross-multiplying gives:
\[
2 \cdot \frac{1}{b(c+a)} = \frac{1}{a(b+c)} + \frac{1}{c(a+b)}
\]
This can be rewritten as:
\[
2c(a+b) = a(b+c) + b(c+a)
\]
### Step 4: Expand and simplify
Expanding both sides:
\[
2c(a+b) = 2ab + ac + bc
\]
This simplifies to:
\[
2ac + 2bc = 2ab + ac + bc
\]
Rearranging gives:
\[
2ac + 2bc - ac - bc = 2ab
\]
Which simplifies to:
\[
ac + bc = 2ab
\]
### Step 5: Rearranging the equation
Rearranging gives:
\[
\frac{1}{a} + \frac{1}{c} = 2 \cdot \frac{1}{b}
\]
This shows that \(a, b, c\) are in H.P. since it satisfies the condition for H.P.
### Conclusion
Thus, we conclude that if \(\frac{1}{a(b+c)}, \frac{1}{b(c+a)}, \frac{1}{c(a+b)}\) are in H.P., then \(a, b, c\) must also be in H.P.
