If a+b+c not equal to zero and b+c/a,c+a/b,a+b/c are in AP.Prove that 1/a,1/b,1/c are also in AP

Show that a^2,b^2,c^2 are in AP if and only iff 1/b+c,1/c+a,1/a+b are in AP

{Experts can you please clarify my doubts}

Given that , b+ca, c+ab, a+bc are in AP.c+ab-b+ca=a+bc-c+abac+a2-b2-bcab=ab+b2-c2-acbcca-b+a-ba+bab=ab-c+b-cb+ccba-ba+b+cab=a+b+cb-cbcSInce a+b+c0, soa-bab=b-cbcNow separating terms we get, 1b-1a=1c-1bor 2b=1c+1aMeans 1a, 1b, 1c are in AP.2)Taking LHS and then proving the same, 1b+c, 1c+a, 1a+b are in AP.So, 2×1a+c=1a+b+1b+c2b+ca+b=b+ca+c+a+ba+c2ab+2b2+2ac+2bc=ac+bc+a2+ab+bc+c2+ab+ac 2ab+2b2+2ac+2bc=2ac+2bc+2ab+a2+c22b2=a2+c2means a2, b2 , c2 are in AP.

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