`"If " a^(2), b^(2), c^(2)" are in A.P., prove that "(1)/(b+c),(1)/(c+a),(1)/(a+b) " are also in A.P."`

Explanation

To prove that if \( a^2, b^2, c^2 \) are in arithmetic progression (A.P.), then \( \frac{1}{b+c}, \frac{1}{c+a}, \frac{1}{a+b} \) are also in A.P., we can follow these steps:

### Step 1: Understand the condition of A.P.
Since \( a^2, b^2, c^2 \) are in A.P., we have:
\[ 2b^2 = a^2 + c^2 \]

### Step 2: Rewrite the terms we want to prove are in A.P.
We need to show that:
\[ \frac{1}{b+c}, \frac{1}{c+a}, \frac{1}{a+b} \]
are in A.P. This means we need to prove:
\[ 2 \cdot \frac{1}{c+a} = \frac{1}{b+c} + \frac{1}{a+b} \]

### Step 3: Find a common denominator for the right-hand side.
The common denominator for \( \frac{1}{b+c} + \frac{1}{a+b} \) is \( (b+c)(a+b) \). Thus, we can write:
\[ \frac{1}{b+c} + \frac{1}{a+b} = \frac{(a+b) + (b+c)}{(b+c)(a+b)} = \frac{a + 2b + c}{(b+c)(a+b)} \]

### Step 4: Evaluate the left-hand side.
Now, we need to evaluate \( 2 \cdot \frac{1}{c+a} \):
\[ 2 \cdot \frac{1}{c+a} = \frac{2}{c+a} \]
To express this with a common denominator, we can write:
\[ \frac{2(b+c)(a+b)}{(c+a)(b+c)(a+b)} \]

### Step 5: Set up the equation.
Now, we need to show:
\[ \frac{2(b+c)(a+b)}{(c+a)(b+c)(a+b)} = \frac{a + 2b + c}{(b+c)(a+b)} \]
Cross-multiplying gives:
\[ 2(b+c)(a+b) = (a + 2b + c)(c + a) \]

### Step 6: Expand both sides.
Expanding the left-hand side:
\[ 2(ab + ac + b^2 + bc) \]
Expanding the right-hand side:
\[ ac + a^2 + 2bc + 2ab + bc + c^2 \]

### Step 7: Simplify the equation.
We need to simplify both sides and check if they are equal. After simplification, if both sides are equal, we conclude that:
\[ \frac{1}{b+c}, \frac{1}{c+a}, \frac{1}{a+b} \text{ are in A.P.} \]

### Conclusion:
Thus, we have shown that if \( a^2, b^2, c^2 \) are in A.P., then \( \frac{1}{b+c}, \frac{1}{c+a}, \frac{1}{a+b} \) are also in A.P.

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