Question

if a²,b²,c² are in A.P, then prove that 1/b+c, 1/c+a, 1/a+b are in A.P

Answer 1

In AP form 2nd term - 1st term = 3rd term - 2nd term
b²-a² = c²-b²
b²+b² = c²+a²
2b² = c²+a²

Add 2ab+2ac+2bc on both sides

2b²+2ab+2ac+2bc = a²+c²+ac+ac+bc+bc+ab+ab
2b²+2ab+2ac+2bc = ac+bc+a²+ab+bc+c²+ab+ac
2b²+2ab+2ca+2cb = ca+cb+a²+ab+cb+c²+ab+ac
2(ba+b²+ca+cb) = (ca+cb+a²+ab) + (cb+c²+ab+ac)
2((ba+b²)+(ca+cb)) = ((ca+cb)+(a²+ab)) + ((cb+c²)+(ab+ac))
2(b(a+b)+c(a+b)) = (c(a+b)+a(a+b)) + (c(b+c)+a(b+c)) 
2(b+c)(a+b) = (c+a)(a+b) + (c+a)(b+c)

Divide whole by (a+b)(b+c)(c+a)

$\frac{2}{c+a} = \frac{1}{b+c} + \frac{1}{a+b}$

$\frac{1}{c+a} + \frac{1}{c+a} = \frac{1}{b+c} + \frac{1}{a+b}$

$\frac{1}{c+a}- \frac{1}{b+c} = \frac{1}{a+b} - \frac{1}{c+a}$
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2nd term - 1st term = 3rd term - 2nd term
Thus  $\frac{1}{b+c} , \frac{1}{c+a} , \frac{1}{a+b}$ are in AP