Show that a square b square c square are in AP if and only if 1/b + c ,1/c+ a,1/ a + b are in AP​

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)2/c+a = 1/b+c + 1/a+b1/c+a + 1/c+a = 1/b+c + 1/a+b1/c+a - 1/b+c = 1/a+b - 1/c+a2nd term - 1st term = 3rd term - 2nd term Thus 1/b+c, 1/c+a, 1/a+b are in AP.HOPE IT HELPS !!!THANK YOU !!!