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

Answer :;

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

Add2ab+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),
we get,

2/c+a = 1/b+c + 1/a+b

hence proved.

Thanks