If a^2,b^2,c^2 are in AP,then prove that a/b+c,b/c+a,c/a+b are in AP

It is given that a^2 , b^2 and c^2 are in an AP
So they have a common difference
b^2 - a^2 = c^2-b^2
(b - a)(b + a) = (c - b)(c + b)
(b - a) / (b + c) = (c - b) / (b + a)

Let;
(b - a) / (b + c) = (c - b) / (b + a) = K

Now for a/(b + c) , b/(c + a) and c/(a+ b)

b/(c+a) - a/(b+c)
= b(b+c) - a(a+c) / (b+c)(c+a)
= b^2 + bc - a^2 - ac / (b+c)(c+a)
= (b-a)(b+a) + c(b-a)
= (b-a)(b+a+c) / (b+c)(c+a)
= K(a+b+c) / (c+a)

c/(a+b) - b/(c+a)
= c^2 + ac - ab - b^2 /
= (c-b)(c+b) + a(c-b)
= (c - b)(a+b+c) / (a+b)(c+a)
= K(a+b+c) / (c+a)

So;

b/(c+a) - a/(b+c) = c/(a+b) - b/(c+a)
Which means that terms a/(b+c) , b/(c+a) and c/(a+b) have a a common difference
Therefore a/(b+c) , b/(c+a) and c/(a+b) are in an A.P